chapter 6 differential equations - seneca valley … 6 differential equations section 6.1 slope...

© 2010 Brooks/Cole, Cengage Learning

C H A P T E R 6 Differential Equations

Section 6.1 Slope Fields and Euler’s Method.......................................................593

Section 6.2 Differential Equations: Growth and Decay.......................................605

Section 6.3 Separation of Variables and Logistic Equation.................................616

Section 6.4 First-Order Linear Differential Equations .........................................629

Review Exercises ........................................................................................................639

Problem Solving .........................................................................................................649

© 2010 Brooks/Cole, Cengage Learning 593

C H A P T E R 6 Differential Equations

Section 6.1 Slope Fields and Euler’s Method

1. Differential equation: 4y y′ =

Solution: 4xy Ce=

Check: 44 4xy Ce y′ = =

2. Differential Equation: 23 5 xy y e−′ + = −

Solution: 2

22

x

x

y e

y e

−

−

=

′ = −

Check: ( ) ( )2 2 23 2 5x x xe e e− − −− + = −

3. Differential equation: 2 22xyy

x y′ =

−

Solution: 2 2x y Cy+ =

Check:

( )

( )

2

2 2 2

2 2

2 2

2 22

2

22

22

2

2

x yy Cyxy

y C

xyyy Cy

xyy x y

xyy x

xyx y

′ ′+ =

−′ =−

−′ =−

−=

− +

−=

−

=−

4. Differential Equation: 2 1dy xydx y

=−

Solution: 2 22 lny y x− =

Check:

2

22 2

1

1

1

yy y xy

y y xy

xyy

yxyy

y

′ ′− =

⎛ ⎞′− =⎜ ⎟

⎝ ⎠

′ =−

′ =−

5. Differential Equation: 0y y′′ + =

Solution: 1 2

1 2

1 2

sin coscos sin

sin cos

y C x C xy C x C xy C x C x

= −

′ = +′′ = − +

Check: ( ) ( )1 2 1 2sin cos sin cos 0y y C x C x C x C x′′ + = − + + − =

6. Differential equation: 2 2 0y y y′′ ′+ + =

Solution: 1 2cos sinx xy C e x C e x− −= +

Check: ( ) ( )

( ) ( )( ) ( )( ) ( )

1 2 1 2

1 2

1 2

1 2 1 2 1 2

1 1 2 2 2 1 2 1

sin cos

2 sin 2 cos

2 2 2 sin 2 cos

2 sin cos 2 cos sin

2 2 2 2 sin 2 2 2 2 cos 0

x x

x x

x x

x x x x

x x

y C C e x C C e x

y C e x C e x

y y y C e x C e x

C C e x C C e x C e x C e x

C C C C e x C C C C e x

− −

− −

− −

− − − −

− −

′ = − + + − +

′′ = −

′′ ′+ + = − +

− + + − + + +

= − − + + − − + + =

594 Chapter 6 Differential Equations


7. Differential Equation: tany y x′′ + =

Solution:

( ) ( )( ) ( )( )

( ) ( )( )( )

2

2

cos ln sec tan

1cos sec tan sec sin ln sec tansec tan

cossec tan sec sin ln sec tan

sec tan

1 sin ln sec tan

1sin sec tan sec cos ln sec tansec tan

sin sec cos ln sec

y x x x

y x x x x x x xx x

xx x x x x x

x x

x x x

y x x x x x x xx x

x x x x

= − +

′ = − ⋅ + + ++

−= + + +

+

= − + +

′′ = ⋅ + + ++

= + + tan x

Check: ( )( )sin sec cos ln sec tan cos ln sec tan tan .y y x x x x x x x x x′′ + = + + − + =

8. Differential Equation: 4 2 xy y e′′ ′+ =

Solution: ( )

( )

4

4 4

4

252 8 245 5 532 25 5

x x

x x x x

x x

y e e

y e e e e

y e e

−

− −

−

= +

′ = − + = − +

′′ = +

Check: 4 432 2 8 2 2 84 4 25 5 5 5 5 5

x x x x x xy y e e e e e e− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞′′ ′+ = + + − + = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

9. 2

2 2

2

sin cos cos

sin cos 2 cos sin

1 2 cos sin 2

y x x x

y x x x x

x x

= −

′ = − + +

= − + +

Differential Equation:

( ) ( )2 22 2 sin cos cos 1 2 cos sin 2

2 sin cos 1 sin 22 sin 2 1

y y x x x x x

x x xx

′+ = − + − + +

= − +

= −

Initial condition , 04π⎛ ⎞⎜ ⎟⎝ ⎠

:

2

2 2 2 2sin cos cos 04 4 4 2 2 2π π π ⎛ ⎞

− = ⋅ − =⎜ ⎟⎜ ⎟⎝ ⎠

10. 212 2 cos 3

2 sin

y x x

y x x

= − −

′ = +

Differential equation: 2 siny x x′ = +

Initial condition ( )0, 5 :− 0 2 cos 0 3 5− − = −

11.

( )

26

2 26 6

4

4 12 48

x

x x

y e

y e x xe

−

− −

=

′ = − = −

Differential equation:

( )2 26 612 12 4 48x xy xy x e xe− −′ = − = − = −

Initial condition ( )0, 4 : 04 4e =

12.

( )

cos

cos cossin sin

x

x x

y e

y e x x e

−

− −

=

′ = = ⋅

Differential Equation:

( )cossin sin sinxy x e x y y x−′ = ⋅ = =

Initial condition , 1 :2π⎛ ⎞⎜ ⎟⎝ ⎠

( )cos 2 0 1e eπ− = =

Section 6.1 Slope Fields and Euler's Method 595


In Exercises 13–20, the differential equation is ( )4 16 = 0.y y−

13. ( )

( )

4

4

3 cos

3 cos

16 45 cos 0,

y x

y x

y y x

=

=

− = − ≠

No

14. ( )

( ) ( )

4

4

2 sin

2 sin

16 2 sin 16 2 sin 0

y x

y x

y y x x

=

=

− = − ≠

No

15. ( )

( )

4

4

3 cos 2

48 cos 2

16 48 cos 2 48 cos 2 0,

y x

y x

y y x x

=

=

− = − =

Yes

16. ( )

( ) ( )

4

4

3 sin 2

48 sin 2

16 48 sin 2 16 3 sin 2 0

y x

y x

y y x x

=

=

− = − =

Yes

17. ( )

( )

2

4 2

4 2 2

16

16 16 16 0,

x

x

x x

y e

y e

y y e e

−

−

− −

=

=

− = − =

Yes

18.

( )

( )

44

44

5 ln30

3016 80 ln 0,

y x

yx

y y xx

=

= −

− = − − ≠

No

19. ( )

( )

2 21 2 3 4

4 2 21 2 3 4

4

sin 2 cos 2

16 16 16 sin 2 16 cos 2

16 0,

x x

x x

y C e C e C x C x

y C e C e C x C x

y y

−

−

= + + +

= + + +

− =

Yes

20. ( )

( ) ( ) ( )

2

4 2

4 2 2

3 4 sin 2

48 64 sin 2

16 48 64 sin 2 16 3 4 sin 2 0,

x

x

x x

y e x

y e x

y y e x e x

= −

= −

− = − − − =

Yes

In Exercises 21–28, the differential equation is 32 = .xxy y x e′ −

21.

( ) ( )2

2 3

, 2

2 2 2 0 ,x

y x y x

xy y x x x x e

′= =

′ − = − = ≠

No

22.

( )3 2

2 3 3 3

, 3

2 3 2 x

y x y x

xy y x x x x x e

′= =

′ − = − = ≠

No

23. ( )( )( ) ( )

2 2 2

2 2 3

, 2 2

2 2 2 ,

x x x x

x x x

y x e y x e xe e x x

xy y x e x x x e x e

′= = + = +

′ − = + − =

Yes

24. ( ) ( ) ( )2 22 , 2 2x x xy x e y x e x e′= + = + +

2 2 2

3

2 2 4 2 2

,

x x x

x

xy y x x e xe x x e x

x e

′ ⎡ ⎤ ⎡ ⎤− = + + − +⎣ ⎦ ⎣ ⎦

=

Yes

25.

( ) ( ) 3

sin , cos

2 cos 2 sin ,x

y x y x

xy y x x x x e

′= =

′ − = − ≠

No

26.

( ) 3

cos , sin

2 sin 2 cos x

y x y x

xy y x x x x e

′= = −

′ − = − − ≠

No

27.

3

1ln ,

12 2 ln ,x

y x yx

xy y x x x ex

′= =

⎛ ⎞′ − = − ≠⎜ ⎟⎝ ⎠

No

28. 2 2 25 , 2 10x x xy x e x y x e xe x′= − = + −

2 2 2

3

2 2 10 2 5

,

x x x

x

xy y x x e xe x x e x

x e

′ ⎡ ⎤ ⎡ ⎤− = + − − −⎣ ⎦ ⎣ ⎦

=

Yes



29. 2xy Ce−= passes through ( )0, 3 .

03 3Ce C C= = ⇒ =

Particular solution: 23 xy e−=

30. ( )2y x y C+ = passes through ( )0, 2 .

( )2 0 2 4C C+ = ⇒ =

Particular solution: ( )2 4y x y+ =

31. 2 3y Cx= passes through ( )4, 4 .

( ) 1416 64C C= ⇒ =

Particular solution: 2 314y x= or 2 34y x=

32. 2 22x y C− = passes through ( )3, 4 .

( )2 9 16 2C C− = ⇒ =

Particular solution: 2 22 2x y− =

33. Differential equation: 4 0yy x′ − =

General solution: 2 24y x C− =

Particular solutions: 0, Two intersecting lines1, 4, Hyperbolas

CC C

=

= ± = ±

34. Differential equation: 0yy x′ + =

General solution: 2 2x y C+ =

Particular solutions: 0,C = Point 1, 4,C C= = Circles

35. Differential equation: 2 0y y′ + =

General solution: 2xy Ce−=

( ) ( )2 22 2 2 0x xy y C e Ce− −′ + = − + =

Initial condition ( ) 00, 3 : 3 Ce C= =

Particular solution: 23 xy e−=

36. Differential equation: 3 2 0x yy′+ =

General solution: 2 23 2x y C+ =

( )6 4 0

2 3 2 0

3 2 0

x yy

x yy

x yy

′+ =

′+ =

′+ =

Initial condition ( )1, 3 :

( ) ( )2 23 1 2 3 3 18 21 C+ = + = =

Particular solution: 2 23 2 21x y+ =

−3 3

−2

C = 0

2

−3 3

−2

C = 1

2

−3 3

−2

C = −1

2

−3 3

−2

C = 4

2

−3 3

−2

C = −4

2

x1 2

1

2

y



37. Differential equation: 9 0y y′′ + =

General solution: 1 2sin 3 cos 3y C x C x= +

( ) ( )

1 2

1 2

1 2 1 2

3 cos 3 3 sin 3 ,9 sin 3 9 cos 3

9 9 sin 3 9 cos 3 9 sin 3 cos 3 0

y C x C xy C x C x

y y C x C x C x C x

′ = −

′′ = − −

′′ + = − − + + =

Initial conditions , 26π⎛ ⎞⎜ ⎟⎝ ⎠

and 1y′ = when :6

x π=

1 2 1

1 2

1 2 2 2

2 sin cos 22 2

3 cos 3 3 sin 3

11 3 cos 3 sin 32 2 3

C C C

y C x C x

C C C C

π π

π π

⎛ ⎞ ⎛ ⎞= + ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′ = −

⎛ ⎞ ⎛ ⎞= − = − ⇒ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Particular solution: 12 sin 3 cos 33

y x x= −

38. Differential equation: 0xy y′′ ′+ =

General solution: 1 2 lny C C x= +

2 2 2

2 22

1 1,

1 1 0

y C y Cx x

xy y x C Cx x

⎛ ⎞ ⎛ ⎞′ ′′= = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞′′ ′+ = − + =⎜ ⎟⎝ ⎠

Initial conditions ( )2, 0 and 12

y′ = when 2:x =

1 2

2

22 1

0 ln 2

1 1, ln 22 2

C CCyx

C C C

= +

′ =

= ⇒ = = −

Particular solution: ln 2 ln ln2xy x= − + =

39. Differential equation: 2 3 3 0x y xy y′′ ′− + =

General solution: 31 2y C x C x= +

( ) ( ) ( )2

1 2 2

2 2 2 32 1 2 1 2

3 , 6

3 3 6 3 3 3 0

y C C x y C x

x y xy y x C x x C C x C x C x

′ ′′= + =

′′ ′− + = − + + + =

Initial conditions ( )2, 0 and 4y′ = when 2:x =

1 2

21 2

1 2

0 2 8

34 12

C C

y C C xC C

= +

′ = +

= +

1 22 1

1 2

12

4 0, 2

12 4C C

C CC C

+ = ⎫= = −⎬

+ = ⎭

Particular solution: 3122y x x= − +



40. Differential equation: 9 12 4 0y y y′′ ′− + =

General solution: ( )2 31 2

xy e C C x= +

( ) ( )( ) ( )

( )( ) ( )( ) ( )( )

2 3 2 3 2 31 2 2 1 2 2

2 3 2 3 2 31 2 2 2 1 2 2

2 3 2 3 2 31 2 2 1 2 2 1 2

2 2 23 3 3

2 2 2 2 2 2 23 3 3 3 3 3 3

2 2 2 2 23 3 3 3 3

2

9 12 4 9 2 12 4 0

x x x

x x x

x x x

y e C C x C e e C C C x

y e C C C x e C e C C C x

y y y e C C C x e C C C x e C C x

′ = + + = + +

′′ = + + + = + +

′′ ′− + = + + − + + + + =

Initial conditions ( )0, 4 and ( )3, 0 :

( )( )( )( )

21 2

1 1

22 2

43

0 3

4 1 0 4

0 4 3

e C C

C C

e C C

= +

= + ⇒ =

= + ⇒ = −

Particular solution: ( )2 3 434xy e x= −

41. 2

2 3

6

6 2

dy xdx

y x dx x C

=

= = +∫

42.

( )

3

43 2

2 3

32 32 2

dy x xdx

xy x x dx x C

= −

= − = − +∫

43.

( )( )

2

22

2

11 ln 1

1 21 , 2

dy xdx x

xy dx x Cx

u x du x dx

=+

= = + ++

= + =

∫

44.

( )4

ln 44

x

x

xx

x

dy edx e

ey dx e Ce

=+

= = + ++∫

45.

2

2 21

21

2 ln ln

dy xdx x x

y dxx

x x C x x C

−= = −

⎛ ⎞= −⎜ ⎟⎝ ⎠

= − + = − +

∫

46.

( ) ( )( )

2

2 2

2

cos

1cos sin2

, 2

dy x xdx

y x x dx x C

u x du x dx

=

= = +

= =

∫

47.

( )

sin 2

1sin 2 cos 22

2 , 2

dy xdx

y x dx x C

u x du dx

=

= = − +

= =

∫

48.

( )

2 2

2

tan sec 1

sec 1 tan

dy x xdxy x dx x x C

= = −

= − = − +∫

49. 6dy x xdx

= −

Let 6,u x= − then 2 6x u= + and 2 .dx u du=

y = ( )( )( )

( )

( ) ( )

( ) ( )

( ) ( )

2

4 2

53

5 2 3 2

3 2

3 2

6 6 2

2 6

2 25

2 6 4 652 6 6 1052 6 45

x x dx u u u du

u u du

u u C

x x C

x x C

x x C

− = +

= +

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

= − + − +

= − − + +

= − + +

∫ ∫∫



50. 2 3dy x xdx

= −

Let 3 ,u x= − then 23x u= − and 2dx u du= −

( ) ( )

( )

( ) ( )

( ) ( )

( ) ( )

2

4 2

53

5 2 3 2

3 2

3 2

2 3 2 3 2

4 12

4 45

4 3 4 354 3 3 55

4 3 25

y x x dx u u u du

u u du

u u C

x x C

x x C

x x C

= − = − −

= −

= − +

= − − − +

= − − − +

= − − + +

∫ ∫∫

51. 2

2 212

x

x x

dy xedx

y xe dx e C

=

= = +∫

( )2 , 2u x du x dx= =

52. 25 xdy edx

−=

( )2 2 215 5 2 102

x x xy e dx e dx e C− − −⎛ ⎞= = − − = − +⎜ ⎟⎝ ⎠∫ ∫

53.

54.

55.

56.

57. sin 2dy xdx

=

For 0, 0.dyxdx

= = Matches (b).

58. 1 cos2

dy xdx

=

For 10, .2

dyxdx

= = Matches (c).

59. 2xdy edx

−=

As , 0.dyxdx

→ ∞ → Matches (d).

60. 1dydx x

=

For 0, dyxdx

= is undefined (vertical tangent). Matches (a).

61. (a), (b)

(c) As ,x y→ ∞ → −∞

As ,x y→ −∞ → −∞

62. (a), (b)

(c) As ,x y→ ∞ → ∞

As ,x y→ −∞ → −∞

x –4 –2 0 2 4 8

y 2 0 4 4 6 8

dy dx –4 Undef. 0 1 43 2

x –4 –2 0 2 4 8

y 2 0 4 4 6 8

dy dx 6 2 4 2 2 0

x –4 –2 0 2 4 8

y 2 0 4 4 6 8

dy dx 2 2− –2 0 0 2 2− –8

x –4 –2 0 2 4 8

y 2 0 4 4 6 8

dy dx 3 0 3− 3− 0 3

8

5

y

x

(4, 2)

−2

4

−4

4

x

y



63. (a), (b)

(c) As ,x y→ ∞ → −∞

As ,x y→ −∞ → −∞

64. (a), (b)

(c) As ,x y→ ∞ → −∞

As ,x y→ −∞ → −∞

65. (a) ( )1, 1, 0yx

′ =

As ,x y→ ∞ → ∞

[Note: The solution is ln .y x= ]

(b) ( )1, 2, 1yx

′ = −

As ,x y→ ∞ → ∞

66. (a) ( )1 , 0, 1yy

′ =

As ,x y→ ∞ → ∞

(b) ( )1 , 1, 1yy

′ =

As ,x y→ ∞ → ∞

67. ( )0.25 , 0 4dy y ydx

= =

(a), (b)

68. ( )4 , 0 6dy y ydx

= − =

(a), (b)

69. ( ) ( )0.02 10 , 0 2dy y y ydx

= − =

(a), (b)

4

−3

5

x

y(2, 2)

−4

y

x

2

−2

−4

−6

2−2−4

(0, −4)

6−1

−2

−3

1

2

3

x

y

(1, 0)

6−1

−2

−3

1

2

3

x

y

(2, −1)

y

x

(0, 1)

3

3

−3

y

x

(1, 1)

3

3

−3

−6 6

−4

12

−50

10

5

−12 48

−2

12



70. ( ) ( )0.2 2 , 0 9dy x y ydx

= − =

(a), (b)

71. ( ) ( )0.4 3 , 0 1dy y x ydx

= − =

(a), (b)

72. ( )81 sin , 0 22 4

xdy ye ydx

π−= =

(a), (b)

73. ( )( ) ( )( )( ) ( )( )

1 0 0 0

2 1 1 1

, 0 2, 10, 0.1

, 2 0.1 0 2 2.2

, 2.2 0.1 0.1 2.2 2.43, etc.

y x y y n h

y y hF x y

y y hF x y

′ = + = = =

= + = + + =

= + = + + =

74. ( )( ) ( )( )( ) ( )( )

1 0 0 0

2 1 1 1

, 0 2, 20, 0.05

, 2 0.05 0 2 2.1

, 2.1 0.05 0.05 2.1 2.2075, etc.

y x y y n h

y y hF x y

y y hF x y

′ = + = = =

= + = + + =

= + = + + =

The table shows the values for 0, 2, 4, , 20.n = …

75. ( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

1 0 0 0

2 1 1 1

3 2 , 0 3, 10, 0.05

, 3 0.05 3 0 2 3 2.7

, 2.7 0.05 3 0.05 2 2.7 2.4375, etc.

y x y y n h

y y hF x y

y y hF x y

′ = − = = =

= + = + − =

= + = + + =

n 0 1 2 3 4 5 6 7 8 9 10

nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

ny 2 2.2 2.43 2.693 2.992 3.332 3.715 4.146 4.631 5.174 5.781

n 0 2 4 6 8 10 12 14 16 18 20

nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

ny 2 2.208 2.447 2.720 3.032 3.387 3.788 4.240 4.749 5.320 5.960

n 0 1 2 3 4 5 6 7 8 9 10

nx 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

ny 3 2.7 2.438 2.209 2.010 1.839 1.693 1.569 1.464 1.378 1.308

−5 50

10

8

−2

−2

8

3

−3

−3

5



76. ( ) ( )( ) ( ) ( )( )( )( ) ( ) ( )( )( )

1 0 0 0

2 1 1 1

0.5 3 , 0 1, 5, 0.4

, 1 0.4 0.5 0 3 1 1

, 1 0.4 0.5 0.4 3 1 1.16, etc.

y x y y n h

y y hF x y

y y hF x y

′ = − = = =

= + = + − =

= + = + − =

77. ( )( ) ( ) ( )

( ) ( ) ( )( )

0 11 0 0 0

0.1 1.12 1 1 1

, 0 1, 10, 0.1

, 1 0.1 1.1

, 1.1 0.1 1.2116, etc.

xyy e y n h

y y hF x y e

y y hF x y e

′ = = = =

= + = + =

= + = + ≈

78. ( )( ) ( )( )( ) ( ) ( ) ( )( )

1 0 0 0

2 1 1 1

cos sin , 0 5, 10, 0.1

, 5 0.1 cos 0 sin 5 5.0041

, 5.0041 0.1 cos 0.1 sin 5.0041 5.0078, etc.

y x y y n h

y y hF x y

y y hF x y

′ = + = = =

= + = + + ≈

= + = + + ≈

79. ( ), 3 , 0, 3xdy y y edx

= =

80. ( )22 , 2 4, 0, 2dy x y xdx y

= = +

n 0 1 2 3 4 5

nx 0 0.4 0.8 1.2 1.6 2.0

ny 1 1 1.16 1.454 1.825 2.201

n 0 1 2 3 4 5 6 7 8 9 10

nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

ny 1 1.1 1.212 1.339 1.488 1.670 1.900 2.213 2.684 3.540 5.958

n 0 1 2 3 4 5 6 7 8 9 10

nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

ny 5 5.004 5.008 5.010 5.010 5.007 4.999 4.985 4.965 4.938 4.903

x 0 0.2 0.4 0.6 0.8 1

( )y x (exact) 3 3.6642 4.4755 5.4664 6.6766 8.1548

( ) ( )0.2y x h = 3 3.6000 4.3200 5.1840 6.2208 7.4650

( ) ( )0.1y x h = 3 3.6300 4.3923 5.3147 6.4308 7.7812

x 0 0.2 0.4 0.6 0.8 1

( )y x (exact) 2 2.0199 2.0785 2.1726 2.2978 2.4495

( ) ( )0.2y x h = 2 2.000 2.0400 2.1184 2.2317 2.3751

( ) ( )0.1y x h = 2 2.0100 2.0595 2.1460 2.2655 2.4131



81. ( ) ( )1cos , sin cos , 0, 02

xdy y x y x x edx

= + = − +

82. As h increases (from 0.1 to 0.2), the error increases.

83. ( ) ( )1 72 , 0, 140 , 0.12

dy y hdt

= − − =

(a)

(b) 272 68 exactty e−= +

(c) ( ) ( )1 72 , 0, 140 , 0.052

dy y hdt

= − − =

The approximations are better using 0.05.h =

84. When 0, 0,x y′= = therefore (d) is not possible.

When , 0, 0x y y′> < (decreasing function) therefore (c) is the equation.

85. The general solution is a family of curves that satisfies the differential equation. A particular solution is one member of the family that satisfies given conditions.

86. A slope field for the differential equation ( ),y F x y′ = consists of small line segments at various

points ( ),x y in the plane. The line segment equals the

slope ( ),y F x y′ = of the solution y at the point ( ), .x y

87. Consider ( ) ( )0 0, , .y F x y y x y′ = = Begin with a point

( )0 0,x y that satisfies the initial condition, ( )0 0.y x y=

Then, using a step size of h, find the point ( ) ( )( )1 1 0 0 0 0, , , .x y x h y hF x y= + + Continue

generating the sequence of points ( ) ( )( )1 1, , , .n n n n n nx y x h y hF x y+ + = + +

88. kx

kx

y Cedy Ckedx

=

=

Because 0.07 ,dy dx y= you have 0.07 .kx kxCke Ce=

So, 0.07.k =

C cannot be determined.

89. False. Consider Example 2. 3y x= is a solution to

3 0,xy y′ − = but 3 1y x= + is not a solution.

90. True

91. True

92. False. The slope field could represent many different differential equations, such as 2 4 .y x y′ = +

t 0 1 2 3

Euler 140 112.7 96.4 86.6

x 0 0.2 0.4 0.6 0.8 1

( )y x (exact) 0 0.2200 0.4801 0.7807 1.1231 1.5097

( ) ( )0.2y x h = 0 0.2000 0.4360 0.7074 1.0140 1.3561

( ) ( )0.1y x h = 0 0.2095 0.4568 0.7418 1.0649 1.4273

t 0 1 2 3

Exact 140 113.24 97.016 87.173

t 0 1 2 3

Euler 140 112.98 96.7 86.9



93. ( ) 22 , 0 4, 4 xdy y y y edx

−= − = =

(a)

(b) If h is halved, then the error is approximately halved ( )0.5 .r ≈

(c) When 0.05,h = the errors will again be approximately halved.

94. ( ), 0 1, 1 2 xdy x y y y x edx

−= − = = − +

(a)

(b) If h is halved, then the error is halved ( )0.5 .r ≈

(c) When 0.05,h = the error will again be approximately halved.

95. (a) ( )

( )

4 12 24

1 24 12 6 34

dIL RI E tdt

dI Idt

dI I Idt

+ =

+ =

= − = −

(b) As , 2.t I→ ∞ → That is, ( )lim 2.t

I t→∞

= In fact,

2I = is a solution to the differential equation.

96.

( )

2

2

2

16 0

16 0

16 0 because 0

4

kt

kt

kt

kt kt

kt

y e

y ke

y k ey y

k e e

k e

k

=

′ =

′′ =

′′ − =

− =

− = ≠

= ±

97.

2

sincos

sin16 0

y A ty A t

y A ty y

ωω ω

ω ω

=

′ =

′′ = −

′′ + =

2

2

sin 16 sin 0

sin 16 0

A t A t

A t

ω ω ω

ω ω

− + =

⎡ ⎤− =⎣ ⎦

If 0,A ≠ then 4ω = ±

x 0 0.2 0.4 0.6 0.8 1

y 4 2.6813 1.7973 1.2048 0.8076 0.5413

1y 4 2.5600 1.6384 1.0486 0.6711 0.4295

2y 4 2.4000 1.4400 0.8640 0.5184 0.3110

1e 0 0.1213 0.1589 0.1562 0.1365 0.1118

2e 0 0.2813 0.3573 0.3408 0.2892 0.2303

r 0.4312 0.4447 0.4583 0.4720 0.4855

x 0 0.2 0.4 0.6 0.8 1

y 1 0.8375 0.7406 0.6976 0.6987 0.7358

1y 1 0.8200 0.7122 0.6629 0.6609 0.6974

2y 1 0.8000 0.6800 0.6240 0.6192 0.6554

1e 0 0.0175 0.0284 0.0347 0.0378 0.0384

2e 0 0.0375 0.0606 0.0736 0.0795 0.0804

r 0.47 0.47 0.47 0.48 0.48

−3 3

−3

3

t

I

Section 6.2 Differential Equations: Growth and Decay 605


98. ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

22 2

, 0

2 2 2

2

f x f x xg x f x g x

f x f x f x f x xg x f x

d f x f x x g x f xdx

′′ ′+ = − ≥

′ ′ ′′ ′+ = − ⎡ ⎤⎣ ⎦

⎡ ⎤′ ′+ = − ⎡ ⎤⎣ ⎦⎣ ⎦

For ( ) ( ) 20, 2 0x x g x f x′< − ⎡ ⎤ ≥⎣ ⎦

For ( ) ( ) 20, 2 0x x g x f x′> − ⎡ ⎤ ≤⎣ ⎦

So, ( ) ( )2 2f x f x′+ is increasing for 0x < and decreasing for 0.x >

( ) ( )2 2f x f x′+ has a maximum at 0.x = So, it is bounded by its value at ( ) ( )2 20, 0 0 .x f f ′= + So, f (and f ′ ) is bounded.

99. Let the vertical line x k= cut the graph of the solution ( )y f x= at ( ), .k t The tangent line at ( ),k t is

( )( )y t f k x k′− = −

Because ( ) ( ),y p x y q x′ + = you have

( ) ( ) ( )y t q k p k t x k− = ⎡ − ⎤ −⎣ ⎦

For any value of t, this line passes through the point ( )

( )( )

1 , .q k

kp k p k

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠

To see this, note that

( )( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )

( )

?

?

1

.

q kt q k p k t k k

p k p k

q k q kq k k p k tk t kq k p k kt t

p k p k

⎛ ⎞− = ⎡ − ⎤ + −⎜ ⎟⎣ ⎦⎜ ⎟

⎝ ⎠

= − + − − + = −

Section 6.2 Differential Equations: Growth and Decay

1.

( )2

3

3 32

dy xdx

xy x dx x C

= +

= + = + +∫

2.

( )2

6

6 62

dy xdx

xy x dx x C

= −

= − = − +∫

3.

1

1

3

31

3

ln 3

3

3

x C x

x

dy ydx

dy dxy

dy dxy

y x C

y e Ce

y Ce

+

= +

=+

=+

+ = +

+ = =

= −

∫ ∫

4.

1

1

6

61

6

ln 6

6

6

x C x

x

dy ydx

dy dxy

dy dxy

y dy x C

y e Ce

y Ce

− + −

−

= −

=−

−= −

−

− = − +

− = =

= −

∫ ∫

5.

2 21

2 2

5

5

5

5

1 52 25

xyy

yy x

yy dx x dx

y dy x dx

y x C

y x C

′ =

′ =

′ =

=

= +

− =

∫ ∫∫ ∫



6.

23 2

1

2 3 2

7

7

7

7 22 3

21 4

xyy

yy x

yy dx x dx

y x C

y x C

′ =

′ =

′ =

= +

− =

∫ ∫

7.

( )

( )

( )

3 21

3 22 3 1

3 22 31

3 22 3

2ln3

x C

xC

x

y x yy xy

y dx x dxy

dy x dxy

y x C

y e

e e

Ce

+

′ =

′=

′=

=

= +

=

=

=

∫ ∫

∫ ∫

8. ( )

( )

( )

2

1

2 2 1

2 21

2 2

1

1

1

1

ln 12

1

1

1

x C

C x

x

y x y

y xy

y dx x dxydy x dx

y

xy C

y e

y e e

Ce

+

′ = +

′=

+′

=+

=+

+ = +

+ =

= −

= −

∫ ∫

∫ ∫

9. ( )

( )( )

( )( )

2

2

2

2

2

21

2

2

2

1 2 0

21

21

21

21

ln ln 1

ln ln 1 ln

ln ln 1

1

x y xy

xyyx

y xy x

y xdx dxy x

dy x dxy x

y x C

y x C

y C x

y C x

′+ − =

′ =+

′=

+′

=+

=+

= + +

= + +

⎡ ⎤= +⎣ ⎦

= +

∫ ∫

∫ ∫

10.

( )

( )

( )

( )

2

1

2

1

2 2 1

2 21

2 2

100

100 100

100

1001

100

ln 1002

ln 1002

100

100

100

x C

xC

x

xy y x

y x xy x y

y xy

y dx x dxy

dy x dxy

xy C

xy C

y e

y e e

y Ce

− −

−−

−

′+ =

′ = + = −

′=

−′

=−

=−

− − = +

− = − −

− =

− = −

= −

∫ ∫

∫ ∫

11. 2

2

dQ kdt t

dQ kdt dtdt t

kdQ CtkQ Ct

=

=

= − +

= − +

∫ ∫

∫

12. ( )

( )

( )

( )

2

2

25

25

252

252

dP k tdt

dP dt k t dtdt

kdP t C

kP t C

= −

= −

= − − +

= − − +

∫ ∫

∫



13. ( )

( )

( )

( )

2

2

500

500

5002

5002

dN k sds

dN ds k s dsds

kdN s C

kN s C

= −

= −

= − − +

= − − +

∫ ∫

∫

14. ( )

( )

( )

2

1

2

1

2 2 1

2 21

2 2

1

1

12

ln2

kx C

kxC

kx

dy kx L ydxdy kx

L y dxdy dx kx dx

L y dx

kxdy CL y

kxL y C

L y e

y L e e

y L Ce

− −

−−

−

= −

=−

=−

= +−

− − = +

− =

− = − +

= −

∫ ∫

∫

15. (a)

(b) ( ) ( )

2

2 22 21

2 21

6 , 0, 0

6

ln 62

6

6

x C x

x

dy x ydx

dy x dxy

xy C

y e C e

y C e

− + −

−

= −

= −−

−− = +

− = =

= +

(0, 0): 1 1

2 2

0 6 6

6 6 x

C C

y e−= + ⇒ = −

= −

16. (a)

(b)

2

2 22 21

01 1

2 2

1, 0,2

ln2

1 1 10, :2 2 2

12

x C x

x

dy xydxdy x dxy

xy C

y e C e

C e C

y e

+

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

= +

= =

⎛ ⎞ = ⇒ =⎜ ⎟⎝ ⎠

=

17. ( )

( )

2

2

2

1 , 0, 102

12

14110 0 1041 104

dy tdt

dy t dt

y t C

C C

y t

=

=

= +

= + ⇒ =

= +

∫ ∫

18. ( )

( )

3 2

3 2

3 2

3 , 0, 104

34

12110 0 1021 102

dy tdt

dy t dt

y t C

C C

y t

= −

= −

= − +

= − + ⇒ =

= − +

∫ ∫

x−5 −1

9

5

y

(0, 0)

−6 6

−1

7

x

)(0, 12

y

−4

−4

4

4

4

−1

−4

(0, 10)

16

0 10

−5

(0, 10)

15



19. ( )

( )

1

2 2 21 1

0

2

1 , 0, 102

12

1ln2

10 10

10

t C C t t

t

dy ydtdy dty

y t C

y e e e Ce

Ce C

y e

− + − −

−

= −

= −

= − +

= = =

= ⇒ =

=

∫ ∫

20. ( )

( )

( )

1

3 4 1

3 4 3 41

0

3 4

3 , 0, 104

34

3ln4

10 10

10

t C

tC t

t

dy ydtdy dty

y t C

y e

e e Ce

Ce C

y e

+

=

=

= +

=

= =

= ⇒ =

=

∫ ∫

21.

( )Theorem 6.1kx

dy kydx

y Ce

=

=

( ) ( )

( ) ( )

0

4

0, 6 : 6

1 54, 15 : 15 6 ln4 2

k

k

Ce C

e k

= =

⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠

( )1 4 ln 5 2 0.22916 6x xy e e⎡ ⎤⎣ ⎦= ≈

When

( )( ) ( )21 4 ln 5 2 8 ln 5 2

8,

25 756 6 64 2

x

y e e

=

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

22.

( )Theorem 6.1kt

dN kNdtN Ce

=

=

( )

( )

0, 250 : 250

400 81, 400 : 400 250 ln ln250 5

k

C

e k

=

= ⇒ = =

( )ln 8 5 0.4700250 250t tN e e= ≈

When ( ) ( )44 ln 8 5 ln 8 5

4

4, 250 250

8 8192250 .5 5

t N e e= = =

⎛ ⎞= =⎜ ⎟⎝ ⎠

23.

( )Theorem 6.1kt

dV kVdtV Ce

=

=

( )

( ) 4

0, 20,000 : 20,000

1 54, 12,500 : 12,500 20,000 ln4 8

k

C

e k

=

⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠

( )1 4 ln 5 8 0.117520,000 20,000t tV e e⎡ ⎤ −⎣ ⎦= ≈

When ( ) ( )( ) ( )3 21 4 ln 5 8 6 ln 5 8

3 2

6, 20,000 20,000

520,000 9882.118.8

t V e e= = =

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

24.

( )Theorem 6.1kt

dP kPdtP Ce

=

=

( )

( )

0, 5000 : 5000

191, 4750 : 4750 5000 ln20

k

C

e k

=

⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠

( )ln 19 20 0.05135000 5000t tP e e−= ≈

When ( )( )ln 19 20 5

5

5, 5000

195000 3868.905.20

t P e= =

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

25. ( )

( ) ( )

5

ln 10 5 5 0.4605

1, 0, , 5, 52

1212152ln 10

51 1 110 or2 2 2

kt

kt

k

t t t

y Ce

C

y e

e

k

y e y e⎡ ⎤⎣ ⎦

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

=

=

=

= = ≈

10

−1

−1

(0, 10)

16

−5 5

−5

(0, 10)

40



26. ( )

( )

5

0.4159

1, 0, 4 , 5,2

4

41 42

ln 1 80.4159

54

kt

kt

k

t

y Ce

C

y e

e

k

y e−

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

=

=

= ≈ −

=

27. ( ) ( )

5

, 1, 5 , 5, 2

5 10 2

2 10 5

kt

k k

k k

y Ce

Ce Ce

Ce Ce

=

= ⇒ =

= ⇒ =

5

5

4

1 4

2 5

2 525

1 2 2ln ln4 5 5

k k

k k

k

Ce Ce

e e

e

k

=

=

=

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )

( )

1 4 1 41 4 ln 2 5

1 41 4 ln 2 5 0.2291

2 55 5 5 55 2

55 6.28722

k

t t

C e e

y e e

−−−

⎡ ⎤ −⎣ ⎦

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

28. ( )

3 3

4 4

1, 3, , 4, 52

1 1 22

15 15

kt

k k

k k

y Ce

Ce Ce

Ce Ce

⎛ ⎞= ⎜ ⎟⎝ ⎠

= ⇒ =

= ⇒ =

3 4

3 4

125

10

10ln 10 2.3026

k k

k k

k

Ce Ce

e e

ek

=

=

=

= ≈

( )

.3026

2.3026 4

2.3026

50.0005

0.0005

t

t

y Ce

CeC

y e

2=

=

≈

=

29. In the model ,kty Ce C= represents the initial value of y (when 0t = ). k is the proportionality constant.

30. dyy kydt

′ = =

31. 12

dy xydx

=

0dydx

> when 0.xy > Quadrants I and III.

32. 212

dy x ydx

=

0dydx

> when 0.y > Quadrants I and II.

33. Because the initial quantity is 20 grams,

20 .kty e=

Because the half-life is 1599 years,

( )

( )1599

1 11599 2

10 20

ln .

ke

k

=

=

So, ( )ln 1 2 159920 .ty e⎡ ⎤⎣ ⎦=

When ( ) ( )ln 1 2 1599 10001000, 20 12.96 .t y e g⎡ ⎤⎣ ⎦= = ≈

When 10,000, 0.26g.t y= ≈

34. Because the half-life is 1599 years,

( )

( )

159912

1 11599 2

1

ln .

ke

k

=

=

Because there are 1.5 g after 1000 years,

( ) ( )ln 1 2 1599 10001.52.314.Ce

C

⎡ ⎤⎣ ⎦=

≈

So, the initial quantity is approximately 2.314 g.

When ( ) ( )ln 1 2 1599 10,00010,000, 2.3140.03 g.

t y e⎡ ⎤⎣ ⎦= =

≈


( )

( )

159912

1 11599 2

1

ln .

ke

k

=

=

Because there are 0.1 gram after 10,000 years,

( ) ( )ln 1 2 1599 10,0000.17.63.Ce

C

⎡ ⎤⎣ ⎦=

≈


When ( ) ( )ln 1 2 1599 10001000, 7.634.95 g.

t y e⎡ ⎤⎣ ⎦= =

≈




( )

( )

571512

1 15715 2

1

ln .

ke

k

=

=

Because there are 3 grams after 10,000 years,

( ) ( )ln 1 2 5715 10,000310.089.Ce

C

⎡ ⎤⎣ ⎦=

≈


When ( ) ( )ln 1 2 5715 10001000, 10.098.94 g.

t y e⎡ ⎤⎣ ⎦= =

≈

37. Because the initial quantity is 5 grams, 5.C =

Because the half-life is 5715 years,

( )

( )5715

1 15715 2

2.5 5

ln .

ke

k

=

=

When ( ) ( )ln 1 2 5715 10001000 years, 5 4.43 g.t y e⎡ ⎤⎣ ⎦= = ≈

When ( ) ( )ln 1 2 5715 10,00010,000 years, 51.49 g.

t y e⎡ ⎤⎣ ⎦= =

≈


( )

( )

571512

1 15715 2

1

ln .

ke

k

=

=

Because there are 1.6 grams when 1000t = years,

( ) ( )ln 1 2 5715 10001.61.806.Ce

C

⎡ ⎤⎣ ⎦=

≈


When ( ) ( )ln 1 2 5715 10,00010,000, 1.8060.54 g.

t y e⎡ ⎤⎣ ⎦= =

≈

39. Because the half-life is 24,100 years,

( )

( )

24,10012

1 124,100 2

1

ln .

ke

k

=

=

Because there are 2.1 grams after 1000 years,

( ) ( )ln 1 2 24,100 10002.12.161.Ce

C

⎡ ⎤⎣ ⎦=

≈


When ( ) ( )ln 1 2 24,100 10,00010,000, 2.1611.62 g.

t y e⎡ ⎤⎣ ⎦= =

≈

40. Because the half-life is 24,100 years,

( )

( )

24,10012

1 124,100 2

1

ln .

ke

k

=

=

Because there are 0.4 grams after 10,000 years,

( ) ( )ln 1 2 24,100 10,0000.40.533.Ce

C

⎡ ⎤⎣ ⎦=

≈


When ( ) ( )ln 1 2 24,100 10001000, 0.5330.52 g.

t y e⎡ ⎤⎣ ⎦= =

≈

41. ( )

( )

159912

1 11599 2ln

kt

k

y Ce

C Ce

k

=

=

=

When ( ) ( )ln 1 2 1599 100100,0.9576 C

t y Ce⎡ ⎤⎣ ⎦= =

≈

Therefore, 95.76% remains after 100 years.

42.

( )

( )

( )

5715

[ln 1 2 5715]

12

1 1ln5715 2

0.15

1ln2ln 0.15

571515,641.8 years

kt

k

t

y Ce

C Ce

k

C Ce

t

t

=

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

⎛ ⎞⎜ ⎟⎝ ⎠=

≈

43. Because 0.064000 ,tA e= the time to double is given by

0.06

0.06

8000 4000

2ln 2 0.06

ln 2 11.55 years.0.06

t

t

e

et

t

=

=

=

= ≈

Amount after 10 years: ( )( )0.06 104000 $7288.48A e= ≈

44. Because 0.05518,000 ,tA e= the time to double is given by

0.055

0.055

36,000 18,000

2ln 2 0.055

ln 2 12.6 years.0.055

t

t

e

et

t

=

=

=

= ≈

Amount after 10 years: ( )( )0.055 1018,000 $31,198.55A e= ≈



45. Because 750 rtA e= and 1500A = when 7.75,t = you have the following.

7.751500 750ln 2 0.0894 8.94%7.75

re

r

=

= ≈ =

Amount after 10 years: ( )0.0894 10750 $1833.67A e= ≈

46. Because 12,500 rtA e= and 25,000A = when 5,t = you have the following.

525,000 12,500ln 2 0.1386 13.86%

5

re

r

=

= ≈ =

Amount after 10 years: ( ) ( )ln 2 5 1012,500 $50,000A e⎡ ⎤⎣ ⎦= =

47. Because 500 rtA e= and 1292.85A = when 10,t = you have the following.

( )

101292.85 500

ln 1292.85 5000.0950 9.50%

10

re

r

=

= ≈ =

The time to double is given by

0.09501000 500ln 2 7.30 years.

0.095

te

t

=

= ≈

48. Because 2000 rtA e= and 5436.56A = when 10,t = you have the following.

( )

105436.56 2000

ln 5436.56 20000.10 10%

10

re

r

=

= ≈ =

The time to double is given by

0.104000 2000ln 2 6.93 years.0.10

te

t

=

= ≈

49. ( )( )12 20

240

0.0751,000,000 112

0.0751,000,000 112

$224,174.18

P

P−

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

≈

50. ( )( )

( )

12 40

480

0.061,000,000 112

1,000,000 1.005 $91,262.08

P

P −

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈

51. ( )( )12 35

420

0.081,000,000 112

0.081,000,000 112

$61,377.75

P

P−

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

52. ( )( )12 25

300

0.091,000,000 112

0.091,000,000 112

$106,287.83

P

P−

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

≈

53. (a) ( )2000 1000 1 0.07

2 1.07ln 2 ln 1.07

ln 2 10.24 yearsln 1.07

t

t

t

t

= +

=

=

= ≈

(b)

( )( )

12

12

0.072000 1000 112

0.0072 112

0.07ln 2 12 ln 112

ln 2 9.93 years12 ln 1 0.07 12

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈+

(c)

( )( )

365

365

0.072000 1000 1365

0.072 1365

0.07ln 2 365 ln 1365

ln 2 9.90 years365 ln 1 0.07 365

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈+

(d) ( )0.07

0.07

2000 1000

2ln 2 0.07

ln 2 9.90 years0.07

t

t

e

et

t

=

=

=

= ≈



54. (a) ( )2000 1000 1 0.6

2 1.06ln 2 ln 1.06

ln 2 11.90 yearsln 1.06

t

t

t

t

= +

=

=

= ≈

(b) 12

12

0.062000 1000 112

0.062 112

0.06ln 2 12 ln 112

1 ln 2 11.58 years0.0612 ln 112

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈⎛ ⎞+⎜ ⎟⎝ ⎠

(c) 365

365

0.062000 1000 1365

0.062 1365

0.06ln 2 365 ln 1365

1 ln 2 11.55 years0.06365 ln 1365

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈⎛ ⎞+⎜ ⎟⎝ ⎠

(d) 0.06

0.06

2000 1000

2ln 2 0.06

ln 2 11.55 years0.06

t

t

e

et

t

=

=

=

= ≈

55. (a) ( )2000 1000 1 0.085

2 1.085ln 2 ln 1.085

ln 2 8.50 yearsln 1.085

t

t

t

t

= +

=

=

= ≈

(b) 12

12

0.0852000 1000 112

0.0852 112

0.085ln 2 12 ln 112

1 ln 2 8.18 years0.08512 ln 1

12

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈⎛ ⎞+⎜ ⎟⎝ ⎠

(c) 365

365

0.0852000 1000 1365

0.0852 1365

0.085ln 2 365 ln 1365

1 ln 2 8.16 years0.085365 ln 1365

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈⎛ ⎞+⎜ ⎟⎝ ⎠

(d) 0.085

0.085

2000 1000

2ln 2 0.085

ln 2 8.15 years0.085

t

t

e

et

t

=

=

=

= ≈

56. (a) ( )2000 1000 1 0.055

2 1.055ln 2 ln 1.055

ln 2 12.95 yearsln 1.055

t

t

t

t

= +

=

=

= ≈

(b) 12

12

0.0552000 1000 112

0.0552 112

0.055ln 2 12 ln 112

1 ln 2 12.63 years0.05512 ln 1

12

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈⎛ ⎞+⎜ ⎟⎝ ⎠

(c) 365

365

0.0552000 1000 1365

0.0552 1365

0.055ln 2 365 ln 1365

1 ln 2 12.60 years0.055365 ln 1365

t

t

t

t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

= ≈⎛ ⎞+⎜ ⎟⎝ ⎠

(d) 0.055

0.055

2000 1000

2ln 2 0.055

ln 2 12.60 years0.055

t

t

e

et

t

=

=

=

= ≈



57. (a)

( ) ( )

0.006

0.006 7

0.006

7 2.3 2.40

2.40

kt t

t

P Ce Ce

P Ce C

P e

−

−

−

= =

= = ⇒ ≈

=

(b) For ( )0.006 1515, 2.40 2.19 milliont P e−= = ≈

(c) Because 0,k < the population is decreasing.

58. (a)

( ) ( )

0.017

0.017 7

0.017

7 80.3 71.29

71.29

kt t

t

P Ce Ce

P Ce C

P e

= =

= = ⇒ ≈

=

(b) For ( )0.017 1515, 71.29 92.0 milliont P e= = ≈

(c) Because 0,k > the population is increasing.

59. (a)

( ) ( )

0.024

0.024 7

0.024

7 6.7 5.66

5.66

kt t

t

P Ce Ce

P Ce C

P e

= =

= = ⇒ ≈

=

(b) For ( )0.024 1515, 5.66 8.11 milliont P e= = ≈


60. (a)

( ) ( )

0.003

0.003 7

0.003

7 10.0 10.21

10.21

kt t

t

P Ce Ce

P Ce C

P e

−

−

−

= =

= = ⇒ ≈

=

(b) For ( )0.003 1515, 10.21 9.76 milliont P e−= = ≈

(c) Because 0,k < the population is decreasing.

61. (a)

( ) ( )

0.036

0.036 7

0.036

7 30.3 23.55

23.55

kt t

t

P Ce Ce

P Ce C

P e

= =

= = ⇒ ≈

=

(b) For ( )0.036 1515, 23.55 40.41 milliont P e= = ≈


62. (a) Because the population increases by a constant each month, the rate of change from month to month will always be the same. So, the slope is constant, and the model is linear. (b) Although the percentage increase is constant each

month, the rate of growth is not constant. The rate of change of y is given by

dy rydt

=

which is an exponential model.

63. (a) ( )100.1596 1.2455 tN =

(b) 400N = when 6.3t = hours (graphing utility)

Analytically,

( )400 100.1596 1.2455

4001.2455 3.9936100.1596

ln 1.2455 ln 3.9936ln 3.9936 6.3 hours.ln 1.2455

t

t

t

t

=

= =

=

= ≈

64. (a) Let .kty Ce=

At time 2: ( )2 2125 125k kCe C e−= ⇒ =

At time 4: ( ) ( )( )4 2 4

2145

145

1 142 5

350 350 125

2 ln

ln 0.5148

k k k

k

Ce e e

e

k

k

−= ⇒ =

=

=

= ≈

( ) ( )

( )

2

2 1 2 ln 14 5

5 62514 14

125

125

125 44.64

kC e

e

−

−

=

=

= = ≈

Approximately 45 bacteria at time 0.

(b) ( ) ( )1 2 ln 14 5 0.514862514 44.64t ty e e= ≈

(c) When 8,t = ( ) ( ) ( )41 2 ln 14 5 8625 625 14

14 14 5 2744.y e= = =

(d) ( ) ( )1 2 ln 14 56251425,000 12.29 hourste t= ⇒ ≈

65. (a) ( )

( )

20

20

19 30 1

30 11

ln 11 300.0502

20

k

k

e

e

k

= −

=

= ≈ −

( )0.050230 1 tN e−≈ −

(b) ( )0.0502

0.0502

25 30 1

16

ln 6 36 days0.0502

t

t

e

e

t

−

−

= −

=

−= ≈

−



66. (a) ( )

( )

30

30

20 30 1

30 10

ln 1 3 ln 3 0.036630 30

k

k

e

e

k

= −

=

−= = ≈ −

( )0.036630 1 tN e−≈ −

(b) ( )0.0366

0.0366

25 30 1

16

ln 6 49 days0.0366

t

t

e

e

t

−

−

= −

=

−= ≈

−

67. (a)

( )1

10 205181

1811205 181 ln 0.01245

10

kt kt

k

P Ce e

e k

= =

= ⇒ = ≈

( )0.012451 181 181 1.01253 ttP e≈ ≈

(b) Using a graphing utility, ( )2 182.3248 1.01091 tP ≈

(c)

The model 2P fits the data better.

(d) Using the model 2,P

( )

( )

( )( )

320 182.3248 1.01091

320 1.01091182.3248

ln 320 182.3248ln 1.01091

51.8 years, or 2011.

t

t

t

=

=

=

≈

68. (a) ( ) 0.0573

4 3 2

1654.2353 1.0590 1654.2353

0.01942 1.3690 21.970 93.66 435.6

t tR e

I t t t t

= =

= − + − +

(b)

According to the model, ( ) 0.057394.79 tR t e′ ≈

(c)

(d) ( ) IP tR

=

69. ( ) 1610 0

010 log , 10II I

Iβ −= =

(a) ( )14

1410 16

1010 10 log 20 decibels10

β−

−−= =

(c) ( )6.5

6.510 16


β−

−−= =

(b) ( )9

910 16


β−

−−= =

(d) ( )4

410 16


β−

−−= =

00

4000

11

00

800

11

00

0.3

11

150500

P1P2

300



70. ( )

( )

10 1016

6.710

10 1016

810

93 10 log 10 log 1610

6.7 log 10

80 10 log 10 log 1610

8 log 10

I I

I II I

I I

−

−

−

−

= = +

− = ⇒ =

= = +

− = ⇒ =

Percentage decrease: ( )6.7 8

6.710 10 100 95%

10

− −

−

⎛ ⎞−≈⎜ ⎟

⎝ ⎠

71. ( ) ( ) 0.10

0.8 0.10

0.8 0.10

0.8 0.10

100,000

100,000

0.4100,000 0.10 0 when 16.

t

t t

t t

t t

A t V t e

e e

e

dA edt t

−

−

−

−

=

=

=

⎛ ⎞= − =⎜ ⎟⎝ ⎠

The timber should be harvested in the year 2024, ( )2008 16 .+ Note: You could also use a graphing utility

to graph ( )A t and find the maximum of ( ).A t Use the

viewing rectangle 0 30x≤ ≤ and 0 600,000.y≤ ≤

72. 0 ln 10ln ln ln 0, 10ln 10 ln 10

R RI I IR I e− −= = = =

(a) 0

8.3

ln ln8.3ln 10

10 199,526,231.5

I I

I

−=

= ≈

(b)

( ) ( )

0

2 22 ln 10 2 ln 10 ln 10

ln ln2ln 10

10R R R R

I IR

I e e e

−=

= = = =

Increases by a factor of 2 ln 10Re or 10 .R

(c)

( )

0ln lnln 101

ln 10

I IR

dRdI I

−=

=

73. Because ( )80dy k ydt

= −

( )

180

ln 80 .

dy k dty

y kt C

=−

− = +

∫ ∫

When 0, 1500.t y= = So, ln 1420.C =

When 1, 1120.t y= = So,

( ) ( )1 ln 1420 ln 1120 80

104ln 1040 ln 1420 ln .142

k

k

+ = −

= − =

So, ( )ln 104 1421420 80.ty e⎡ ⎤⎣ ⎦= +

When 5, 379.2 F.t y= ≈ °

74. ( )

( )( )

( )

( ) ( )

( )

0

5

5

1 5 ln 2 7

ln 2 7

20

20 See Example 6

160 20 140

60 20 14027

1 2ln 0.250555 7

30 20 140

1 214 71 2ln ln

14 5 715 ln 5 ln 1414 10.53 minutes2 7ln ln

7 2

kt

k

k

k

t

t st s

dy k ydt

y Ce

Ce C

e

e

k

e

e

t

t

= −

= +

= + ⇒ =

= +

=

⎛ ⎞= ≈ −⎜ ⎟⎝ ⎠

= +

⎛ ⎞= = ⎜ ⎟⎝ ⎠

=

= = ≈

It will take 10.53 5 5.53 minutes− = longer.

75. False. If , constant.kt kty Ce y Cke′= = ≠

76. True

77. True

78. True



Section 6.3 Separation of Variables and the Logistic Equation

1.

2 2

1

2 2

2 2

dy xdx y

y dy x dx

y x C

y x C

=

=

= +

− =

∫ ∫

2. 2

2

2 2

33

1

3 3

3

3

33

dy xdx y

y dy x dx

y x C

y x C

=

=

= +

− =

∫ ∫

3. 2

2

2

2 3

1

2 3

5 0

5

5

52 3

15 2

dyx ydxdyy xdx

y dy x dx

y x C

y x C

+ =

= −

= −

−= +

+ =

∫ ∫

4.

( )

2

2

2 2

33

1

3 3

36

6 3

2 33

6 9

dy xdx y

y dy x dx

xy x C

y x x C

−=

= −

= − +

− + =

∫ ∫

5.

1

0.75 1

0.75

0.75

0.75

ln 0.75s C

s

dr rdsdr dsrr s C

r e

r Ce

+

=

=

= +

=

=

∫ ∫

6.

2

2

0.75

0.75

0.752

0.375

dr sdsdr s ds

sr C

r s C

=

=

= +

= +

∫ ∫

7. ( )

( )

( )

3

3

2 3

32

ln 3 ln 2 ln ln 2

2

x y y

dy dxy x

y x C C x

y C x

′+ =

=+

= + + = +

= +

∫ ∫

8.

ln ln ln ln

xy ydy dxy xy x C Cxy Cx

′ =

=

= + =

=

∫ ∫

9.

2

1

2

4 sin

4 sin

4 sin

4 cos2

8 cos

yy xdyy xdx

y dy x dx

y x C

y C x

′ =

=

=

= − +

= −

∫ ∫

10. ( )

( )

( )( )

( )

2

2

8 cos

8 cos

8 cos

8 sin2

16 sin

yy x

dyy xdx

y dy x dx

xy C

y x C

π

π

π

ππ

ππ

′ = −

= −

= −

−= +

−= +

∫ ∫

11.

( ) ( )

2

2

2

1 22

2

1 4

1 4

1 41 1 4 881 1 44

x y xxdy dx

xxdy dx

x

x x dx

y x C

−

′− =

=−

=−

= − − −

= − − +

∫ ∫

∫

12. 2

2

2

2

16 1111

1611

16

11 16

x y xdy xdx x

xdy dxx

y x C

′− =

=−

=−

= − +

∫ ∫

Section 6.3 Separation of Variables and the Logistic Equation 617


13.

( )

( )( ) ( )

21

2 21 2 ln ln 21

ln 0

ln ln ,

1ln ln2

x C x

y x xy

dy x dxdx u x duy x x

y x C

y e Ce+

′− =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= +

= =

∫ ∫

14.

2

12 7 0

12 7

12 7

6 7

x

x

x

x

yy edyy edx

y dy e dx

y e C

′ − =

=

=

= +

∫ ∫

15.

2

2 0

2

2

22

x

x

x

x

yy edyy edx

y dy e dx

y e C

′ − =

=

=

= +

∫ ∫

Initial condition ( )0, 3 : 9 522 2

C C= + ⇒ =

Particular solution: 2

2

522 2

4 5

x

x

y e

y e

= +

= +

16. 1 2 1 2

3 2 3 21

3 2 3 2

0

2 23 3

x y y

y dy x dx

y x C

y x C

′+ =

= −

= − +

+ =

∫ ∫

Initial condition ( )1, 9 :

( ) ( )3 2 3 29 1 27 1 28 C+ = + = =

Particular solution: 3 2 3 2 28y x+ =

17. ( )

( )

( )

( )

2

1

21 2

1 0

1

1ln

2x

y x y

dy x dxy

xy C

y Ce− +

′+ + =

= − +

+= − +

=

∫ ∫

Initial condition ( ) 1 2 1 22, 1 : 1 ,Ce C e−− = =

Particular solution: ( ) ( )2 21 1 2 2 2x x x

y e e⎡ ⎤− + − +⎢ ⎥⎣ ⎦= =

18.

( )

2

2

2 ln 0

2 2 ln

ln

ln2

xy xdyx xdx

xdy dxx

xy C

′ − =

=

=

= +

∫ ∫

Initial condition ( )1, 2 : 2 C=

Particular solution: ( )21 ln 22

y x= +

19. ( ) ( )

( ) ( )

( ) ( ) ( )( )

2 2

2 2

2 21

2 2 2

2 2

1 1

1 11 1ln 1 ln 12 2

ln 1 ln 1 ln ln 1

1 1

y x y x y

y xdy dxy x

y x C

y x C C x

y C x

′+ = +

=+ +

+ = + +

⎡ ⎤+ = + + = +⎣ ⎦

+ = +

Initial condition ( )0, 3 : 1 3 4C C+ = ⇒ =

Particular solution: ( )2 2

2 2

1 4 1

3 4

y x

y x

+ = +

= +

20.

( ) ( )( ) ( )

2 2

1 2 1 22 2

1 2 1 22 2

1 1

1 1

1 1

dyy x x ydx

y y dy x x dx

y x C

− −

− = −

− = −

− − = − − +

∫ ∫

Initial condition ( )0, 1 : 0 1 1C C= − + ⇒ =

Particular solution: 2 21 1 1y x− = − −

21.

( )

2

2

21

2cos 2

sin

sin

1ln cos2

v

du uv vdvdu v v dvu

u v C

u Ce−

=

=

= − +

=

∫ ∫

Initial condition: ( ) 1 21 210 1:u C e

e−= = =

Particular solution: ( )21 cos 2vu e

−=



22. 2

2

212

r s

r s

r s

dr eds

e dr e ds

e e C

−

− −

− −

=

=

− = − +

∫ ∫

Initial condition:

( ) 1 10 0: 12 2

r C C= − = − + ⇒ = −


2

22

2

1 12 2

1 12 2

1 1 1ln ln2 2 2

2ln1

r s

r s

ss

s

e e

e e

er e

re

− −

− −

−−

−

− = − −

= +

⎛ ⎞+⎛ ⎞− = + = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= ⎜ ⎟+⎝ ⎠

23.

1

0

lnkt

dP kP dtdP k dtPP kt C

P Ce

− =

=

= +

=

∫ ∫

Initial condition: ( ) 00 00 ,P P P Ce C= = =

Particular solution: 0ktP P e=

24. ( )

( ) 1

70 0

70ln 70

70 kt

dT k T dt

dT k dtTT kt C

T Ce−

+ − =

= −−− = − +

− =

∫ ∫

Initial condition: ( ) 00 140: 140 70 70T Ce C= − = = =

Particular solution: ( )70 70 , 70 1kt ktT e T e− −− = = +

25.

22

4

4

22

dy xydx y

y dy x dx

xy C

′ = =

=

= +

∫ ∫

Initial condition ( ) ( )20, 2 : 2 2 0 8C C= + ⇒ =


2

2 2

2 82

4 16

xy

y x

= +

− =

26.

2 2

916

16 9

982

dy xdx y

y dy x dx

y x C

−=

= −

−= +

∫ ∫

Initial condition ( ) 9 251, 1 : 8 ,2 2

C C= − + =

Particular solution: 2 2

2 2

9 2582 2

16 9 25

y x

y x

−= +

+ =

27.

1

2

22 1

2 ln ln ln ln

dy yydx x

dy dxy x

y x C x C

y Cx

′ = =

=

= + = +

=

∫ ∫

Initial condition ( ) 19, 1 : 1 99

C C= ⇒ =


2

19

9 013

y x

y x

y x

=

− =

=

28.

3 2

3 2

23

3 2

ln ln ln

dy ydx x

dy dxy x

y x C

y Cx

=

=

= +

=

∫ ∫

Initial condition ( ) ( )3 2 18, 2 : 2 8 ,8

C C= =

Particular solution: 3 2 2 318 ,2

y x y x= =

29. ( )

1

2

02 2

12

1ln2

x

dy y ymdx x x

dy dxy

y x C

y Ce−

−= = = −

+ −

= −

= − +

=

∫ ∫



30. 00

dy y ymdx x x

dy dxy x

−= = =

−

=∫ ∫

1ln ln ln ln lny x C x C Cxy Cx= + = + =

=

31. ( )( )

( )

3 2 3

3 3 2 2 3 3

3 3 2 3

, 4

, 4

4

f x y x xy y

f tx ty t x txt y t y

t x xy y

= − +

= − +

= − +

Homogeneous of degree 3

32. ( )( )

3 2 2 2

3 3 4 2 2 2 2

, 3 2

, 3 2

f x y x x y y

f tx ty t x t x y t y

= + −

= + −

Not homogeneous

33. ( )

( )

2 2

2 2

4 2 2 2 23

2 2 2 2 2 2

,

,

x yf x yx y

t x y x yf tx ty tt x t y x y

=+

= =+ +


34. ( )

( )

2 2

2 2 2 2

2

2 2 2 2

,

,

xyf x yx y

tx tyf tx tyt x t y

t xy xytt x y x y

=+

=+

= =+ +


35. ( )( ) [ ]

( )2 2

, 2 ln

, 2 ln

2 ln 2 ln ln

f x y xy

f tx ty txty

t xy t xy

=

=

⎡ ⎤= = +⎣ ⎦

Not homogeneous

36. ( ) ( )( ) ( ) ( )

, tan

, tan tan

f x y x y

f tx ty tx ty t x y

= +

= + = ⎡ + ⎤⎣ ⎦

Not homogeneous

37. ( )

( )

, 2 ln

, 2 ln 2 ln

xf x yytx xf tx tyty y

=

= =


38. ( )

( )

, tan

, tan tan

yf x yxty yf tx tytx x

=

= =


39.

( )

( )

( )

( )

( )

2

2

2

2

2

2

,2

21 1

2 2

21

ln 1 ln ln ln

11

1

1 /

x yy y vxx

dv x vxv xdx xdv v vx vdx

dv dxv x

v x C Cx

Cxv

Cxy x

x Cxx y

x C x y

+′ = =

++ =

+ −= − =

=−

− − = + =

=−

=⎡ − ⎤⎣ ⎦

=−

= −

∫ ∫

40. ( )

( )

( ) ( ) ( )( )

3 3

2

2 3 3

2 33

4 2 3 3 3 3 3

2

,

x yy

xy

xy dy x y dx

y vx dy x dv v dx

x vx x dv v dx x vx dx

x v dv x v dx x dx v x dx

xv dv dx

+′ =

= +

= = +

+ = +

+ = +

=

2

3

3

3 3 3

1

ln3

3 ln

3 ln

v dv dxx

v x C

y x Cx

y x x Cx

=

= +

⎛ ⎞ = +⎜ ⎟⎝ ⎠

= +

∫ ∫



41.

2

2

121

22

2

2 2

2 2

,

111 1 21 1

12 1

1 ln 2 1 ln ln ln2

2 1

2 1

2

x yy y vxx y

dv x xvv xdx x xv

vv dx x dv dxvv v vx dv v dx dxv v

v dxdvv v x

Cv v x Cx

Cv vx

y y Cx x x

y xy x C

−′ = =+

−+ =

+−

+ =+− − −⎛ ⎞= − =⎜ ⎟+ +⎝ ⎠

+= −

+ −

+ − = − + =

+ − =

+ − =

+ − =

∫ ∫

42.

( )

2 2

2 2 2

2

2

2

2

2

2

2

2 2

,2

212 2

21

ln 1 ln ln ln

1

1

x yy y vxxy

dv x v xv xdx x v

vv dx x dv dxv

v dxdvv x

Cv x Cx

Cvx

y Cx x

y x Cx

+′ = =

++ =

++ =

= −−

− = − + =

− =

− =

− =

∫ ∫

43. 2 2

2

2 2 2

2

3

2 2

2

3

1 12

12

2

12

2 22

,

1

1 1

1

1 ln ln ln ln2

1 ln2

ln2

x y

xyy y vxx y

dv x vv xdx x x v

vv dx x dv dxvv vx dv v dx dx

v v

v dxdvv x

v x C C xv

C xvvx C yy

y Ce−

′ = =−

+ =−

+ =−

⎛ ⎞⎛ ⎞= − = ⎜ ⎟⎜ ⎟− −⎝ ⎠ ⎝ ⎠

−=

− − = + =

−=

−=

=

∫ ∫

44.

2 2

2

2

2

3

2 3 ,

2 3 2 3

2 2 21

ln 1 ln ln ln

1

1

1

x yy y vxx

dv x vxv x vdx xdv dv dxx vdx v xv x C x C

v x Cy x Cxy x Cxy Cx x

+′ = =

++ = = +

= + ⇒ =+

+ = + =

+ =

+ =

= −

= −

∫ ∫



45. ( )( ) ( )

/2 0,

2 0

y x

v

x dy xe y dx y vx

x v dx x dv xe vx dx

−

−

− + = =

+ − + =

21

/ 21

/ 2

2

ln

ln ln

ln

v

v

y x

y x

e dv dxx

e C x

e C x

e C x

=

=

= +

= +

∫ ∫

Initial condition ( )1, 0 : 1 C=

Particular solution: / 21 lny xe x= +

46. ( )( )( )

2

2 2 2 2

0,

0

y dx x x y dy y vx

x v dx x x v v dx x dv

− + + = =

− + + + =

11

1

1

1

1

ln ln ln ln

ln

v

y x

y x

v dxdvv x

Cv v x CxCvxv

C evxC ey

y Ce−

+= −

+ = − + =

=

=

=

=

∫ ∫

Initial condition ( ) 11, 1 : 1 Ce C e−= ⇒ =

Particular solution: 1 y xy e −=

47.

( ) ( )( )

sec 0,

sec 0

sec

yx y dx x dy y vxx

x v xv dx x v dx x dv

v v dx v dx x dv

⎛ ⎞+ − = =⎜ ⎟⎝ ⎠

+ − + =

+ = +

( )

1

sin

sin

cos

sin ln lnv

y x

dxv dvx

v x C

x Ce

Ce

=

= +

=

=

∫ ∫

Initial condition ( ) 01, 0 : 1 Ce C= =

Particular solution: ( )sin y xx e=

48. ( )2 22 0x y dx xy dy+ + =

Let , .y vx dy x dv v dx= = +

( ) ( )( )

( )( )

( )

( )( )

( )

( )

2 2 2

2 2 2 3

2

2

21

1 22 2

1 22 2

1 22 1 22 22 2

1 22 2

2 0

2 2 0

2 2

2112 ln ln 12

ln ln 1 ln

1

1 1

1

x v x dx x vx x dv v dx

x x v dx x v dv

v dx xv dv

vdx dvx v

x v C

x v C

x C v

y CC x yx x x

C x yx

−

−

+ + + =

+ + =

+ = −

−=

+

− = + +

= + +

= +

⎛ ⎞= + = +⎜ ⎟

⎝ ⎠

= +

Initial condition ( ) ( )1, 0 : 1 1 0 1C C= + ⇒ =

Particular solution: 2 2

2 2

1

1

x yx

x x y

= +

= +

49. dy xdx

=

212

y x dx x C= = +∫

50. dy xdx y

= −

2 2

1

2 2

2 2

y dy x dx

y x C

y x C

= −

−= +

+ =

4

2

−4

−2x

y

x−4 −2 2 4

−4

−2

2

4

y



51. 4dy ydx

= −

1

1

4

ln 4

4

4

x C

x

dy dxy

y x C

y e

y Ce

− +

−

=−

− = − +

− =

= +

∫ ∫

52. ( )

( ) ( )

( )

21

2 21 8 1 81

21 8

0.25 4

0.254

10.254 4

1ln 48

4

4

C x x

x

dy x ydx

dy x dxy

dy x dx x dxy

y x C

y e Ce

y Ce

− −

−

= −

=−

= − = −−

− = − +

− = =

= +

∫ ∫ ∫

53. (a) Euler's Method gives 0.1602y ≈ when 1.x =

(b)

( )

21

23

23

6

6

ln 3

0 5 5

5

x

x

dy xydxdy xy

y x C

y Ce

y C

y e

−

−

= −

= −

= − +

=

= ⇒ =

=

∫ ∫

(c) At ( )3 11, 5 0.2489.x y e−= = ≈

Error: 0.2489 0.1602 0.0887− ≈


(b)

2

2

21

2

22

6

6

1 3

131 13

31 3

1 9 133

dy xydxdy x dxy

x Cy

yx C

CC

yxx

= −

= −

−= − +

=+

= ⇒ =

= =++

∫ ∫

(c) At ( )

3 31, 0.3.9 1 1 10

x y= = = =+

Error: 0.3 0.2622 0.0378− =


(b) 22 123 4

dy xdx y

+=

−

( ) ( )2

3 2

3 4 2 12

4 12

y dy x dx

y y x x C

− = +

− = + +

∫ ∫

( ) ( )3

3 2

1 2: 2 4 2 1 12 13

4 12 13

y C C

y y x x

= − = + + ⇒ = −

− = + −

(c) At 2,x =

( )

( )( )

3 2

3

2

4 2 12 2 13 15

4 15 0

3 3 5 0 3.

y y

y y

y y y y

− = + − =

− − =

− + + = ⇒ =

Error: 3.0318 3 0.0318− =

56. (a) Euler's Method gives 1.7270y ≈ when 1.5.x =

(b) ( )

( )( )

( )

2

2

2

2

2

2

2 1

21

arctan

arctan 0 1 1

arctan 1

tan 1

dy x ydx

dy x dxy

y x C

C C

y x

y x

= +

=+

= +

= + ⇒ = −

= −

= −

∫ ∫

(c) At ( )21.5, tan 1.5 1 3.0096.x y= = − ≈

Error: 1.7270 3.0096 1.2826− = −

8

4321−3−4x

y

x−4 −2

−2

2

8

4

4

y



57. , ktdy ky y Cedt

= =

Initial amount: ( ) 00y y C= =

Half-life: ( )15990021 1ln

1599 2

ky y e

k

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( )ln 1 2 1599 ty Ce⎡ ⎤⎣ ⎦=

When 50, 0.9786t y C= = or 97.86%.

58. , ktdy ky y Cedt

= =

Initial conditions: ( ) ( )0

0 40, 1 35

40

35 407ln8

k

y y

Ce C

e

k

= =

= =

=

=

Particular solution: ( )ln 7 840 ty e=

When 75% has been changed:

( )

( )

( )( )

ln 7 8

ln 7 8

10 4014

ln 1 410.38 hours

ln 7 8

t

t

e

e

t

=

=

= ≈

59. (a) ( )4dy k ydx

= −

(b) The direction field satisfies ( ) 0dy dx = along

4;y = but not along 0.y = Matches (a).

60. (a) ( )4dy k xdx

= −


4.x = Matches (b).

61. (a) ( )4dy ky ydx

= −


0y = and 4.y = Matches (c).

62. (a) 2dy kydx

=


0,y = and grows more positive as y increases. Matches (d).

63. ( )

1

1

1200

1200ln 1200

1200

1200

kt C kt

kt

dw k wdt

dw k dtww kt C

w e Ce

w Ce

− + −

−

= −

=−− = − +

− = =

= −

∫ ∫

( )0 60 1200 1200 60 1140

1200 1140 kt

w C C

w e−= = − ⇒ = − =

= −

(a)

0.8k = 0.9k = 1k =

(b) 0.8: 1.31 years0.9: 1.16 years1.0: 1.05 years

k tk tk t

= =

= =

= =

(c) Maximum weight: 1200 pounds lim 1200

xw

→∞=

0

1400

1000

1400

1000

1400

100



64. From Exercise 63:

( )

( )0 0

0

1200 , 1

1200

0 1200 1200

1200 1200

kt

t

t

w Ce k

w Ce

w w C C w

w w e

−

−

−

= − =

= −

= = − ⇒ = −

= − −

65. Given family (circles): 2 2

2 2 0x y C

x yyxyy

+ =

′+ =

′ = −

Orthogonal trajectory (lines):

ln ln ln

yyx

dy dxy x

y x K

y Kx

′ =

=

= +

=

∫ ∫

66. Given family (hyperbolas): 2 222 4 0

2

x y Cx yy

xyy

− =

′− =

′ =

Orthogonal trajectory:

22

2

2

ln 2 ln ln

yyx

dy dxy xy x k

ky kxx

−

−′ =

= −

= − +

= =

∫ ∫

67. Given family (parabolas): 2

2

22 2 2

x Cyx Cy

x x yyC x y x

=

′=

′ = = =

Orthogonal trajectory (ellipses):

22

1

2 2

2

2

22

xyy

y dy x dx

xy K

x y K

′ = −

= −

= − +

+ =

∫ ∫

68. Given family (parabolas): 2

2

22 2

12 2

y Cxyy C

C y yyy x y x

=′ =

⎛ ⎞′ = = =⎜ ⎟⎝ ⎠

Orthogonal trajectory (ellipse):

22

1

2 2

2

2

22

xyy

y dy x dx

y x K

x y K

′ = −

= −

= − +

+ =

∫ ∫

69. Given family: 2 3

2

2 2 2

3

2 3

3 3 32 2 2

y Cxyy Cx

Cx x y yyy y x x

=

′ =

⎛ ⎞′ = = =⎜ ⎟

⎝ ⎠

Orthogonal trajectory (ellipses):

22

1

2 2

23

3 2

32

3 2

xyy

y dy x dx

y x K

y x K

′ = −

= −

= − +

+ =

∫ ∫

−6 6

−4

4

−3

−2

3

2

−6 6

−4

4

− 6 6

−4

4

−6 6

−4

4



70. Given family (exponential functions): x

x

y Ce

y Ce y

=

′ = =

Orthogonal trajectory (parabolas):

2

1

2

1

22

yy

y dy dx

y x K

y x K

′ = −

= −

= − +

= − +

∫ ∫

71. 121 xy

e−=

+

Because ( )0 6,y = it matches (c) or (d).

Because (d) approaches its horizontal asymptote slower than (c), it matches (d).

72. 121 3 xy

e−=

+

Because ( ) 120 3,4

y = = it matches (a).

73. 12112

xy

e−=

+

Because ( ) 120 8,32

y = =⎛ ⎞⎜ ⎟⎝ ⎠

it matches (b).

74. 212

1 xye−

=+

Because ( )0 6,y = it matches (c) or (d).

Because y approaches 12L = faster for (c), it matches (c).

75. ( ) 0.752100

1 29 tP te−

=+

(a) 0.75k =

(b) 2100L =

(c) ( ) 21000 701 29

P = =+

(d) 0.75

0.75

0.75

210010501 29

1 29 2129

10.75 ln ln 2929

ln 29 4.4897 yr0.75

t

t

t

ee

e

t

t

−

−

−

=+

+ =

=

⎛ ⎞− = = −⎜ ⎟⎝ ⎠

= ≈

(e) ( )0.75 1 , 0 702100

dP PP Pdt

⎛ ⎞= − =⎜ ⎟⎝ ⎠

76. ( ) 0.25000

1 39 tP te−

=+

(a) 0.2k =

(b) 5000L =

(c) ( ) 50000 1251 39

P = =+

(d) 0.2

0.2

0.2

500025001 39

1 39 2139

10.2 ln ln 3939

ln 39 18.31780.2

t

t

t

ee

e

t

t

−

−

−

=+

+ =

=

⎛ ⎞− = = −⎜ ⎟⎝ ⎠

= ≈

(e) ( )0.2 1 , 0 1255000

dP PP Pdt

⎛ ⎞= − =⎜ ⎟⎝ ⎠

77. 3 1100

dP PPdt

⎛ ⎞= −⎜ ⎟⎝ ⎠

(a) 3k =

(b) 100L =

(c)

−6 6

−4

4

00

120

5



(d) 2

2 3 1 3100 100

3 23 3 1 1 3 1 9 1 1 9 1 1100 100 100 100 100 100 100 100 100

d P P PP Pdt

P P P P P P P P PP P P P

′−⎛ ⎞ ⎛ ⎞′= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − − − − = − − − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦ ⎣ ⎦

2

2 0d Pdt

= for 50,P = and by the first Derivative Test, this is a maximum. 100Note: 502 2LP⎛ ⎞= = =⎜ ⎟

⎝ ⎠

78.

( )

20.1 0.0004

0.1 1 0.004

0.1 1250

dP P Pdt

P P

PP

= −

= −

⎛ ⎞= −⎜ ⎟⎝ ⎠

(a) 10.110

k = =

(b) 250L =

(c)

(d) 250 125.2

P = = (Same argument as in Exercise 77)

79. ( )1 , 0 436

1, 3636

1 1kt t

dy yy ydtk L

Lybe be− −

⎛ ⎞= − =⎜ ⎟⎝ ⎠

= =

= =+ +

( ) 360, 4 : 4 81

bb

= ⇒ =+

Solution: 361 8 ty

e−=

+

80. ( )

( )

2.8

2.8 1 , 0 710

2.8, 1010

1 110 10 30, 7 : 7 1

1 7 7

kt t

dy yy ydtk L

Lybe be

b bb

− −

⎛ ⎞= − =⎜ ⎟⎝ ⎠

= =

= =+ +

= ⇒ + = ⇒ =+

Solution: 2.8

10317

ty

e−=

⎛ ⎞+ ⎜ ⎟⎝ ⎠

81. ( )

( )

2

0.8

4 4 1 , 0 85 150 5 120

4 0.8, 1205

1201 1

1200, 8 : 8 141

kt t

dy y y yy ydt

k L

Lybe be

bb

− −

⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

= = =

= =+ +

= ⇒ =+

Solution: 0.8120

1 14 tye−

=+

82. ( )

( )

( )

2

3 20

3 3 1 , 0 1520 1600 20 240

3 , 24020

2401 1

2400, 15 : 15 151

kt t

dy y y yy ydt

k L

Lybe be

bb

− −

⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

= =

= =+ +

= ⇒ =+

Solution: ( )3 20240

1 15ty

e−

=+

83. (a) ( ), 200, 0 251 kt

LP L Pbe−

= = =+

( )2

2

2

20025 71

200391 72001 7392339

1 23 1 39ln ln 0.26402 39 2 23

k

k

k

bb

e

e

e

k

−

−

−

= ⇒ =+

=+

+ =

=

⎛ ⎞ ⎛ ⎞= − = ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0.2640200

1 7 tPe−

=+

(b) For 5, 70 panthers.t P= ≈

00

300

100



(c) 0.264

0.264

2001001 7

1 7 2

10.264 ln7

7.37 years

t

t

ee

t

t

−

−

=+

+ =

⎛ ⎞− = ⎜ ⎟⎝ ⎠

≈

(d)

( )

1

0.264 1 , 0 25200

dP PkPdt L

PP P

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎝ ⎠

Using Euler's Method, 65.6P ≈ when 5.t =

(e) P is increasing most rapidly where 200 2 100,P = = corresponds to 7.37 years.t ≈

84. (a)

( ) ( )

2

2

2

0.7791

, 20, 0 1, 2 41

201 191

2041 19

1 19 5

19 4

1 4 1 19ln ln 0.77912 19 2 4

201 19

kt

k

k

k

t

Ly L y ybe

bb

ee

e

k

ye

−

−

−

−

−

= = = =+

= ⇒ =+

=+

+ =

=

⎛ ⎞ ⎛ ⎞= − = ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=+

(b) For 5, 14.43 gramst y= ≈

(c) 0.7791

0.7791

0.7791

0.7791

20181 1920 101 1918 911991

1711 1ln 6.60 hours

0.7791 171

t

t

t

t

e

e

e

e

t

−

−

−

−

=+

+ = =

=

=

− ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

(d) 1 191 ln 12 4 20

dy y yky ydt L

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(e) The weight is increasing most rapidly when 2 20 2 10,y L= = = corresponding to 3.78 hours.t ≈

t 0 1 2 3 4 5

Exact 1 2.06 4.00 7.05 10.86 14.43

Euler 1 1.74 2.98 4.95 7.86 11.57



85. A differential equation can be solved by separation of variables if it can be written in the form

( ) ( ) 0.dyM x N ydx

+ =

To solve a separable equation, rewrite as,

( ) ( )M x dx N y dy= −

and integrate both sides.

86. ( ) ( ), , 0,M x y dx N x y dy+ = where M and N are

homogeneous functions of the same degree. See Example 7a.

87. Two families of curves are mutually orthogonal if each curve in the first family intersects each curve in the second family at right angles.

88. Answers will vary. Sample answer: There might be limits on available food or space.

89.

( )( )

( ) ( )

( ) ( )

( ) ( )

2

11

1

1

1 1

1 11 1

11 111

kt

kt

kt

kt

kt kt

kt

kt kt

ktkt

ybe

y bkebe

k bebe be

k bebe be

k ky ybebe

−

−

−

−

− −

−

− −

−−

=+

−′ = −+

= ⋅+ +

+ −= ⋅

+ +

⎛ ⎞= ⋅ − = −⎜ ⎟++ ⎝ ⎠

90. (a) ( )

1lnkt

dv k W vdt

dv k dtW vW v kt C

v W Ce−

= −

=−

− − = +

= −

∫ ∫

Initial conditions: 20, 0 when 0W v t= = = and 10v =

when 0.5t = so, 20, ln 4.C k= =

Particular solution:

( )( )ln 4 120 1 20 14

ttv e−

⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

or

( )1.38620 1 tv e−= −

(b) ( ) ( )1.386 1.38620 1 20 0.7215t ts e dt t e C− −= − ≈ + +∫

Because ( )0 0, 14.43s C= ≈ − and you have

( )1.38620 14.43 1 .ts t e−≈ + −

91. False. dy xdx y

= is separable, but 0y = is not a solution.

92. True

( )( )2 1dy x ydx

= − +

93. False

( )

( )

2 2 2 2 2

2

, 4 6 1

,

f tx ty t x xyt t y

t f x y

= − + +

≠

94. True

2 2 2 22 2x y Cy x y Kxdy x dy K xdx C y dx y

+ = + =

−= =

−

2

2

2

2

2 2 2

2 2 2

2 2

2 2

2 22 2

22

1

x K x Kx xC y y Cy y

Kx xCy y

x y xx y y

y xx y

− −⋅ =

− −

−=

−

+ −=

+ −

−=

−

= −

95.

( )Product Rule

0

0

fg gf f g

f f g gf

fg gf f

′ ′ ′ ′+ =

′ ′ ′− + =

′′ + =

′−

Need ( )2 2 22 1 2 0,x x xf f e xe x e′− = − = − ≠ so

avoid 1.2

x =

( )

( )

( )

2

2

1

1 2

2 112 12 1

1ln ln 2 12

2 1

x

x

x

g f xeg f f xx e

g x x x C

g x Ce x

′ ′= = = +

′ − −−

= + − +

= −

So there exists g and interval ( ), ,a b as long as

( )1 , .2

a b∉

Section 6.4 First-Order Linear Differential Equations 629


Section 6.4 First-Order Linear Differential Equations

1.

( )

3

2 3

11 1 1

x

x

x y xy e

y y ex x

′ + = +

′ + = +

Linear

2.

( ) ( )( )

2 ln

ln 1 2 0

1 20

ln

xy y x y

x y x y

xy y

x

′− =

′ + − =

−′ + =

Linear

3. 2siny y x xy′ − =

Not linear, because of the 2-term.xy

4. 2 5

2 55 2

y xy

y xyy xy

′−=

′− =

′ + =

Linear

5. 1 6 2dy y xdx x

⎛ ⎞+ = +⎜ ⎟⎝ ⎠

Integrating factor: ( )1 lnx dx xe e x∫ = =

( ) 3 2

2

6 2 2

2

xy x x dx x x C

Cy x xx

= + = + +

= + +

∫

6. 2 3 5dy y xdx x

+ = −

Integrating factor: 22 ln 2x dx xe e x∫ = =

( )

32 2 4

22

3 53 54 3

3 54 3

xx y x x dx x C

Cy x xx

= − = + +

= + +

∫

7. 16y y′ − =

Integrating factor: 1dx xe e− −∫ =

16

16 16

16

x x x

x x x

x

e y e y e

ye e dx e C

y Ce

− − −

− − −

′ − =

= = − +

= − +

∫

8. 2 10y xy x′ + =

Integrating factor: 22x dx xe e∫ =

2 2 2

2

10 5

5

x x x

x

ye xe dx e C

y Ce−

= = +

= +

∫

9. ( )( )( )

1 cos

1 cos cos cos

cos cos

y x dx dy

y y x y x x

y x y x

+ =

′ = + = +

′ − =

Integrating factor: cos sinx dx xe e− −∫ =

( ) ( )( )

sin sin sin

sin sin

sin

sin

cos cos

cos

1

x x x

x x

x

x

y e x e y x e

ye x e dx

e C

y Ce

− − −

− −

−

′ − =

=

= − +

= − +

∫

10. ( )( )1 sin 0

sin sin

y x dx dy

y x y x

⎡ − ⎤ − =⎣ ⎦′ − = −

Integrating factor: sin cosx dx xe e−∫ =

cos cos cos

cos

sin

1

x x x

x

ye xe dx e C

y Ce−= − = +

= +

∫

11. ( ) 21 1

1 11

x y y x

y y xx

′− + = −

⎛ ⎞′ + = +⎜ ⎟−⎝ ⎠

Integrating factor: ( )1 1 ln 1 1x dx xe e x−⎡ ⎤ −⎣ ⎦∫ = = −

( ) ( )

( )

2 31

3

11 13

33 1

y x x dx x x C

x x Cyx

− = − = − +

− +=

−

∫

12. 33 xy y e′ + =

Integrating factor: 3 3dx xe e∫ =

3 3 3 6 6

3 3

16

16

x x x x x

x x

ye e e dx e dx e C

y e Ce−

= = = +

= +

∫ ∫



13. 323 xy x y e′ − =

Integrating factor: 2 33x dx xe e− −∫ =

( )

3 3 3

3

x x x

x

ye e e dx dx x C

y x C e

− −= = = +

= +

∫ ∫

14. tan secy y x x′ + =

Integrating factor: tan ln cos secx dx xe e x−∫ = =

2sec sec tan

sin cos

y x x dx x C

y x C x

= = +

= + ⋅∫

15. (a) Answers will vary.

(b)

Integrating factor:

x

dxx x

dy e ydxdy y e e edx

= −

∫+ = =

( )2

2

2

2

x x x

x x

x x

e y e y e

ye e dx

ye e C

′ + =

=

1= +

∫

( ) 1 10 1 12 2

y C C= ⇒ = + ⇒ =

( )

21 12 21 1 12 2 2

x x

x x x x

ye e

y e e e e− −

= +

= + = +

(c)

16. (a)

(b) ( ) ( )

( ) ( )

2 2

1 ln

1 1sin , , sin

x dx x

y y x P x Q x xx x

u x e e x

′ + = = =

∫= = =

2

2 2

2

2

sin1sin cos2

1 1 cos2

1 1 10 cos2 2

1 1 1cos2 2

y x y x x

yx x x dx x C

y x Cx

C C

y xx

ππ

′ + =

= = − +

⎡ ⎤= − +⎢ ⎥⎣ ⎦

⎡ ⎤= − + ⇒ = −⎢ ⎥⎣ ⎦

⎡ ⎤= − −⎢ ⎥⎣ ⎦

∫

(c)

17.

( )2

2 2

cos 1 0

sec sec

y x y

y x y x

′ + − =

′ + =

Integrating factor: 2sec tanx dx xe e∫ =

tan 2 tan tan

tan

sec

1

x x x

x

ye xe dx e C

y Ce−= = +

= +

∫

Initial condition: ( )0 5, 4y C= =

Particular solution: tan1 4 xy e−= +

18. 23 1

213 3

2

2 1

x

x

x y y e

y y ex x

′ + =

⎛ ⎞′ + =⎜ ⎟⎝ ⎠

Integrating factor: ( ) ( )3 22 1x dx xe e

−∫ =

2113 2

2212

1 121

2

x

x

ye dx Cx x

Cxy ex

− = = − +

⎛ ⎞−= ⎜ ⎟

⎝ ⎠

∫

Initial condition: ( )1 , 3y e C= =


23 1

2x xy e

x⎛ ⎞−

= ⎜ ⎟⎝ ⎠

x−4

−3

4

5

y

−6

−2

6

6

4

−4

−4

4

x

y

4

4

−4

−4



19. tan sec cosy y x x x′ + = +

Integrating factor: tan ln sec secx dx xe e x∫ = =

( )sec sec sec cos tan

sin cos cos

y x x x x dx x x C

y x x x C x

= + = + +

= + +∫


Particular solution: ( )sin 1 cosy x x x= + +

20. sec secy y x x′ + =

Integrating factor: sec ln sec tan sec tanx dx x xe e x x+∫ = = +

( ) ( )sec tan sec tan sec

sec tan

1sec tan

y x x x x x dx

x x CCy

x x

+ = +

= + +

= ++

∫

Initial condition: ( )0 4, 4 1 , 31 0

Cy C= = + =+


3 3 cos1 1sec tan 1 sin

xyx x x

= + = ++ +

21. 1 0y yx

⎛ ⎞′ + =⎜ ⎟⎝ ⎠

Integrating factor: ( )1 lnx dx xe e x∫ = =

Separation of variables:

1 1

ln ln lnln ln

dy ydx x

dy dxy x

y x Cxy Cxy C

= −

= −

= − +

=

=

∫ ∫


Particular solution: 4xy =

22. ( )2 1 0y x y′ + − =

Integrating factor: ( ) 22 1x dx x xe e− −∫ =

2

2

x x

x x

ye C

y Ce

−

−

=

=


( )

21

21

2

1 1 2

ln lnx x

x x

dy x dxy

y C x x

yC e

y Ce

−

−

= −

+ = −

=

=

∫ ∫


Particular solution: 22 x xy e −=

23. ( )2

2 21

x dy x y dx

dy x y ydx x x x

= + +

+ += = + +

( ) ( )1

1 21 Linear

1x dx

dy ydx x x

u x ex

−

− = +

∫= =

22 1 1 21

2ln

2 ln

y x dx x dxx x x x

x x Cx

x x Cx

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−⎡ ⎤= + +⎢ ⎥⎣ ⎦= − + +

∫ ∫

( )1 10 2 12

2 ln 12

y C C

y x x x

= = − + ⇒ =

= − + +

24.

( ) ( )

3

2

1 21 2

2

1 1 Linear2 2 2

1x dx

xy y x x

dy xydx x

u x ex

−

′ − = −

− = −

∫= =

2 3 2 1 21 2 1 2

1 2

5 21 2 1 2

3

1 12 2 2 2

5

5

x x xy x dx x dxx

xx x C

x x C x

−⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎡ ⎤= − +⎢ ⎥

⎣ ⎦

= − +

∫ ∫

( )

3

64 174 2 4 25 5

175 5

y C C

xy x x

= = − + ⇒ = −

= − −



25.

( )

( )

1

2

2

3

, constant

1

1 ln

lnkt C

kt

kt

dP kP N Ndt

dP dtkP N

dP dtkP N

kP N t Ck

kP N kt C

kP N e

C e NPk

NP Cek

+

= +

=+

=+

+ = +

+ = +

+ =

−=

= −

∫ ∫

0

0 0

0

When 0:

kt

t P PN NP C C Pk k

N NP P ek k

= =

= − ⇒ = +

⎛ ⎞= + −⎜ ⎟⎝ ⎠

26.

( )

( )

1

2

2

3

1 ln

lnrt C

rt

rt

dA rA Pdt

dA dtrA P

dA dtrA P

rA P t Cr

rA P rt C

rA P e

C e PAr

PA Cer

+

= +

=+

=+

+ = +

+ = +

+ =

−=

= −

∫ ∫

( )

When 0: 0

0

1rt

t AP PC Cr r

PA er

= =

= − ⇒ =

= −

27. (a) ( )( )( )0.08 10

1

275,000 1 $4,212,796.940.06

rtPA er

A e

= −

= − ≈

(b) ( )( )0.059 25550,000 1 $31,424,909.750.05

A e= − ≈

28. ( )

( )

0.08

0.08

125,0001,000,000 10.08

1.64

ln 1.646.18 years

0.08

t

t

e

e

t

= −

=

= ≈

29. (a) , constantdQ q kQ qdt

= −

(b) Q kQ q′ + =

Let ( ) ( ), ,P t k Q t q= = then the integrating factor

is ( ) .ktu t e=

kt kt kt kt ktq qQ e qe dt e e C Cek k

− − −⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∫

0

0 0

0

When 0:

kt

t Q Qq qQ C C Qk kq qQ Q ek k

−

= =

= + ⇒ = −

⎛ ⎞= + −⎜ ⎟⎝ ⎠

(c) limt

qQk→∞

=

30. (a) ( )75dN k Ndt

= −

(b) 75N kN k′ + =

Integrating factor: k dt kte e∫ =

( )75

75

75 75

75

kt kt kt

kt kt

kt kt kt

kt

N e kNe ke

Ne ke

Ne ke e C

N Ce−

′ + =

′ =

= = +

= +

∫

(c) For 1, 20:

20 75 55k k

t N

Ce Ce− −

= =

= + ⇒ − =

For 20 20

20, 35:

35 75 40k k

t N

Ce Ce− −

= =

= + ⇒ − =

19

2055 11 1 11ln40 8 19 8

0.0168

kk

kCe e k

Ce

−

−⎛ ⎞= ⇒ = ⇒ = ⎜ ⎟⎝ ⎠

≈

55

55 55.9296

k

k

Ce

C e

− = −

= − ≈ −

0.016875 55.9296 tN e−= −



31. Let Q be the number of pounds of concentrate in the solution at any time t. Because the number of gallons of solution in the tank at any time t is ( )0 1 2v r r t+ − and

because the tank loses 2r gallons of solution per minute, it must lose concentrate at the rate

( ) 2

0 1 2.Q r

v r r t⎡ ⎤⎢ ⎥

+ −⎢ ⎥⎣ ⎦

The solution gains concentrate at the rate 1 1.r q Therefore, the net rate of change is

( )1 1 2

0 1 2

dQ Qq r rdt v r r t

⎡ ⎤= − ⎢ ⎥

+ −⎢ ⎥⎣ ⎦

or

( )

21 1

0 1 2.dQ r Q q r

dt v r r t+ =

+ −

32. From Exercise 31, and using 1 2 ,r r r= =

10

.dQ rQ q rdt v

+ =

33. (a) ( )

( )

2

1 10 1 2

0 0 1 0

1 2

0 , 25, 0, 200,

110, 10, 020

r QQ q rv r r t

Q q q q v

r r Q Q

′ + =+ −

= = = =

′= = + =

( )

1

1 20

1 120

1ln ln20

t

dQ dtQ

Q t C

Q Ce−

= −

= − +

=

∫ ∫

Initial condition: ( )0 25, 25Q C= =

Particular solution: ( )1 2025 tQ e−=

(b) ( )1 2015 25

3 1ln5 20

320 ln 10.2 min5

te

t

t

−=

⎛ ⎞ = −⎜ ⎟⎝ ⎠

⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

(c) ( )1 20lim 25 0t

te−

→∞=

34. (a) ( )

( )

2

1 10 1 2

0 1 0

1 2

0 25, 0.04, 200,

110, 10, 0.420

r QQ q rv r r t

Q q q v

r r Q Q

′ + =+ −

= = = =

′= = + =

Integrating factor: 1 20te

( )

( )

( )( )

1 201 20 1 20

1 2

1 20

0.4 8

8

0 25 8 17

8 17

tt t

t

t

Qe e dt e C

Q Ce

Q C C

Q e

−

−

−

= = +

= +

= = = + ⇒ =

= +

∫

(b) ( )1 20

1 20

15 8 17

7 17

t

t

e

e

−

−

= +

=

7 1 7ln 20 ln 17.75 min17 20 17

t t⎛ ⎞ ⎛ ⎞= − ⇒ = − ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(c) ( )lim 8 lbst

Q t→∞

=

35. (a) The volume of the solution in the tank is given by ( )0 1 2 .v r r t+ − Therefore, ( )100 5 3 200t+ − = or

50t = minutes.

(b) ( )

( )

21 1

0 1 2

0 0 1 0 1

2

0 , 0, 0.5, 100, 5,

33, 2.5100 2

r QQ q rv r r t

Q q q q v r

r Q Qt

′ + =+ −

= = = = =

′= + =+

Integrating factor: ( ) ( )3 23 100 2 50t dte t+⎡ ⎤⎣ ⎦∫ = +

( ) ( ) ( )

( ) ( )

3 2 3 2 5 2

3 2

50 2.5 50 50

50 50

Q t t dt t C

Q t C t −

+ = + = + +

= + + +

∫

Initial condition:

( ) ( )3 2 5 20 0, 0 50 50 , 50Q C C−= = + = −


( ) ( )

( ) ( )

3 25 2

3 25 2

50 50 50

50 100 50 100

25100 82.32 lbs2

Q t t

Q

−−

−

= + − +

= −

= − ≈



(c) The volume of the solution is given by ( ) ( )0 1 2 100 5 3 200 50v r r t t t+ − = + − = ⇒ =

minutes.

( )

( )

21 1

0 1 2

0 1 0 1 20 0, 1, 100, 5, 3

3 5100 2

r QQ q rv r r t

Q q q v r r

QQt

′ + =+ −

= = = = = =

′ + =+

Integrating factor is ( )3 250 ,t+ as in #43.

( ) ( ) ( )

( ) ( )

3 2 3 2 5 2

3 2

50 5 50 2 50

2 50 50

Q t t dt t C

Q t C t −

+ = + = + +

= + + +

∫

( )

( ) ( ) ( )

( ) ( ) ( )

3 2 3 2 5 2

5 2 3 2

0 0:

0 100 50 100 50 2 50

2 50 2 50 50

Q

C C

Q t t

−

−

=

= + ⇒ = − = −

= + − +

When 50,t =

( ) ( )5 2 3 2 50200 2 50 100 2002

164.64 lbs (double the answer to part (b))

Q −= − = −

≈

36. ( ) ( )y P x y Q x′ + =

Integrating factor: ( )P x dxu e∫=

( ) ( )

( ) ( )

y u P x yu Q x u

uy Q x u

′ + =

′ =

so ( ) ( )u x P x u′ =

Answer (a)

37. From Example 6,

( )1 , Solutionkt m

dv kv gdt m

mgv ek

−

+ =

= −

( ) 8 132, 8, 5 101,4

g mg v mg−

= − = − = − = =

implies that ( )( )5 1 48101 1 .kek

−−− = −

Using a graphing utility, 0.050165,k ≈ and

( )0.2007159.47 1 .tv e−= − −

As , 159.47 ft/sec.t v→ ∞ → − The graph of v is shown below.

38. ( ) ( )

( )

( )( )

0.2007

0.2007

0.2007

159.47 1

159.47 794.57

0 5000 794.57 5794.57

159.47 794.57 5794.57

t

t

t

s t v t dt

e dt

t e C

s C C

s t t e

−

−

−

=

= − −

= − − +

= = − + ⇒ =

= − − +

∫∫

The graph of ( )s t is shown below.

( ) 0s t = when 36.33 sec.t ≈

39. 00,dI R EL RI E I I

dt L L′+ = + =

Integrating factor: ( )R L dt Rt Le e∫ =

0 0

0

Rt L Rt L Rt L

Rt L

E EI e e dt e CL R

EI CeR

−

= = +

= +

∫

40. ( ) 00 0, 120 volts, 600 ohms,I E R= = =

4 henrysL =

( )

( ) ( )

( )( )

0

150

150

150

150

120 10600 51 15 51lim amp5

1 10.90 0.18 15 5

0.9 1

0.1

150 ln 0.1

ln 0.10.0154 sec

150

Rt L

t

t

t

t

t

EI CeR

C C

I e

I

e

e

e

t

t

−

−

→∞

−

−

−

= +

= + ⇒ = −

= −

=

= = −

= −

=

− =

= ≈−

41. ( ) ( )

( ) ( )

Standard form

Integrating factorP x dx

dy P x y Q xdx

u x e

+ =

∫=

50

−200

0 40

0 40

−500

6000



42. ( ) ( ) Standard formny P x y Q x y′ + =

Let ( )1 0, 1 .nz y n−= ≠ Multiplying by

( )1 nn y−− produces

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

11 1 1

1 1 . Linear

n nn y y n P x y n Q x

z n P x z n Q x

− −′− + − = −

′ + − = −

43.

2

2 0

2

y x

dy x dx

y x C

′ − =

=

= +

∫ ∫

Matches c.

44.

1

2

2 0

2

ln 2x

y ydy dxyy x C

y Ce

′ − =

=

= +

=

∫ ∫

Matches d.

45.

21

2

2 0

2

lnx

y xydy x dxy

y x C

y Ce

′ − =

=

= +

=

∫ ∫

Matches a.

46.

( ) 21

22

2

2

2 11 1ln 2 12 2

2 112

x

x

y xy xdy x dx

y

y x C

y C e

y Ce

′ − =

=+

+ = +

+ =

= − +

∫ ∫

Matches b.

47. 2 2 3

2 2

3

3, , 3

y x y x y

n Q x P x

′ + =

= = =

( ) ( ) ( )2 22 3 2 32 2

3 32 2 2 2

3 32 2 2

32 2

322

2

2

1313

1 13

x dx x dx

x x

x x

x

x

y e x e dx

y e x e dx

y e e C

y Ce

Cey

− −−

− − −

− − −

−

∫ ∫= −

= −

= +

= +

+ +

∫∫

48. 1

221, , , x dx x

y xy xy

n Q x P x e e

−′ + =

∫= − = = =

2 2 22

22

2

1

x x x

x

y e xe dx e C

y Ce−

= = +

= +

∫

49.

( )

2

1

1 ln 1

1

2, ,x dx x

y y xyx

n Q x P x

e e x

−

− − −

⎛ ⎞′ + =⎜ ⎟⎝ ⎠

= = =

∫ = =

( )1 1 1

2

2

1

1

y x x x dx x C

x Cxy

yCx x

− − −= − = − +

= − +

=−

∫

50.

( )( ) ( )

1

1 2 1 1 2 ln

1

1, ,2x dx x

y y x yx

n Q x P x

e e x

−

⎛ ⎞′ + =⎜ ⎟⎝ ⎠

= = =

= =

( )

( )

1 2 1 2 1 2

5 25 2

1

25 2

12

15 5

25

y x x x dx

x Cx C

x Cy

x

=

+= + =

+=

∫

51. 3

3

22 ln 2

1

13, 1, ,dx xx

xy y xy

y y yx

n Q P e e xx

−− −

′ + =

′ + =

∫= = = = =

2 2 2 1

2 2

2 22 2

2 2

21 1or 2

2

y x x dx C x C

y x Cx

y x Cxx Cx y

− − − −

−

= − + = +

= +

= = ++

∫

52. ( )

3

2 1 23, 1, 1, dx x

y y y

n P Q e e− −

′ − =

∫= = − = =

( )2 2 2 2

2 2

22

2

11

1

x x x

x

x

y e e dx e C

y Ce

yCe

−

− −

−

= − = − +

= − +

=− +

∫



53.

( ) ( )

3

2 3 2 3

13, , , 1x x

dx x

y y e y n Q e P

e e− −

′ − = = = = −

∫ =

( ) ( ) ( )

( ) ( )

2 3 2 3 1 32 3

2 3 1 32 3

2 3 2 3

2 23 3

2

2

x x xx

x x

x x

y e e e dx e dx

y e e C

y e Ce

− −

−

= =

= +

= +

∫ ∫

54. 2

1

22

x

x

yy y ey y e y−

′ − =

′ − =

1, , 2xn Q e P= − = = −

( )2 2 4

2 4 4 3

2 4

23

23

2

dx x

x x x x

x x

e e

y e e e dx e C

y e Ce

− −

− − −

∫ =

= = − +

= − +

∫

55. (a)

(b) 21dy y xdx x

− =

Integrating factor: 1 ln 1x dx xe ex

− −= =

2

2

3

1 1

12

2

y y xx x

xy x dx Cx

xy Cx

′ − =

⎛ ⎞ = = +⎜ ⎟⎝ ⎠

= +

∫

( ) ( )

( ) ( )

32

32

8 12, 4 : 4 2 4 4 82 2 2

8 12, 8 : 8 2 2 2 42 2 2

xC C y x x x

xC C y x x x

−− = − ⇒ = − ⇒ = − = −

= + ⇒ = ⇒ = + = +

(c)

56. (a)

(b) 3 34y x y x′ + =

Integrating factor: 3 44x dx xe e∫ =

4 4 43 3

4 4 43

4

14

14

4x x x

x x x

x

y e x ye x e

ye x e dx e C

y Ce−

′ + =

= = +

= +

∫

( )( )

4

4

7 7 13 131 12 2 4 4 4 4

3 31 1 1 12 2 4 4 4 4

0, :

0, :

x

x

C C y e

C C y e

−

−

= + ⇒ = ⇒ = +

− − = + ⇒ = − ⇒ = −

(c)

− 4

−6

10

4

−6

10

− 4 4

−1

−1

5

3

−1

−1

5

3



57. (a)

(b) ( )cot 2y x y′ + =

Integrating factor: cot ln sin sinx dx xe e x∫ = =

( )sin cos 2 sin

sin 2 sin 2 cos

2 cot csc

y x x y x

y x x dx x C

y x C x

′ + =

= = − +

= − +∫

( )

( )

1 2 cot 11,1 : 1 2 cot 1 csc 1 sin 1 2 cos 1csc 1

2 cot sin 1 2 cos 1 csc

C C

y x x

+= − + ⇒ = = +

= − + +

( )

( )

2 cot 3 13, 1 : 1 2 cot 3 csc 3 2 cos 3 sin 3csc 3

2 cot 2 cos 3 sin 3 csc

C C

y x x

−− − = − + ⇒ = = −

= − + −

(c)

58. (a)

(b) 22y xy xy′ + =

Bernoulli equation, 2n = letting 1 2 1,z y y− −= = you obtain 22x dx xe e− −= and ( ) 2 211 .

2x xxe dx e− −− =∫ The solution is:

2 21

22

2

12

1 1 1 22 2

2

1 2

x x

xx

x

y e e C

CeCey

yCe

− − −= +

+= + =

=+

( )

( )( )

22

2

2 2 10, 3 : 3 1 21 2 3 6

2 631 3

2 10,1 : 1 1 2 21 2 2

21

xx

x

C CC

yee

C CC

ye

= ⇒ + = ⇒ = −+

= =−−

= ⇒ + = ⇒ =+

=+

(c)

−2

−3

6

3

−2

−3

6

3

−5

−3

7

5

−5

−3

7

5



59. 2 0x y x ye dx e dy+ −− =


2

2

21

2

12

2

x y x y

x y

x y

x y

e e dx e e dy

e dx e dy

e e C

e e C

−

−

−

−

=

=

= − +

+ =

∫ ∫

60. ( )

34

dy xdx y y

−=

+


( ) ( )2

3 22

1

3 2 2

4 3

2 33 2

2 12 3 18

y y dy x dx

y xy x C

y y x x C

+ = −

+ = − +

+ = − +

∫ ∫

61. ( )cos cos 0y x x dx dy− + =


( )( )

sin

1cos1

sin ln 1 ln

ln 1 sin ln

1x

x dx dyy

x y C

y x C

y Ce−

−=

−

= − − +

− = − +

= +

∫ ∫

62. 22 1y x y′ = −


( )

2

2

2

1 21

arcsin

sin

dy x dxy

y x C

y x C

=−

= +

= +

∫ ∫

63. ( ) ( )2 23 4 2 0y xy dx xy x dy+ + + =

Homogeneous: ,y vx dy v dx x dv= = +

( ) ( )( )

( )

2 2 2 2 2

2

5 2

5 2

3 2 4

3 4 2 0

5 2 1 0

ln ln ln

v x vx dx vx x v dx x dv

vdx dvx v v

x v v C

x v v C

x y x y C

+ + + + =

+⎛ ⎞+ =⎜ ⎟+⎝ ⎠

+ + =

+ =

+ =

∫ ∫

64. ( ) 0x y dx x dy+ − =

Linear: 1 1y yx

′ − =

Integrating factor: ( ) 1ln1 1xx dxe ex

−−∫ = =

( )

1 1 ln

ln

y dx x Cx xy x x C

= = +

= +

∫

65. ( )2 0xy e dx x dy− + =

Linear: 2 1 xy y ex x

⎛ ⎞′ + =⎜ ⎟⎝ ⎠

Integrating factor: ( ) 22 ln 2x dx xe e x∫ = =

( )

( )

2 2

2 2

1 1

1

x x

x

yx x e dx e x Cx

e Cy xx x

= = − +

= − +

∫

66. ( )2 2 0y xy dx x dy+ − =

Homogeneous: ,y vx dy v dx x dv= = +

( ) ( )2 2 2 2

2

2

0

01 1

1ln

ln

v x vx dx x v dx x dv

v dx x dv

dx dvx v

x Cv

xyC x

+ − + =

− =

=

= − +

=−

∫ ∫

67. ( )2 4 3 3

3 3

1 0

1

x y dx x y dy

y y x yx

− −

− + =

⎛ ⎞′ + =⎜ ⎟⎝ ⎠

Bernoulli: 3 13, , ,n Q x P x− −= − = =

( ) 44 ln 4x dx xe e x∫ = =

( )( )4 4 3 4 2

4 4 2

4 2

2

y x x x dx x C

x y x C

−= = +

− =

∫

Review Exercises for Chapter 6 639


68. ( )3 4 0y dx x y dy+ + =

Homogeneous: ,x vy dx v dy y dv= = +

( ) ( )

( )( )

4

4

3

3 4 0

1 41

ln 1 ln ln

1

y v dy y dv vy y dy

dv dyv y

v y C

y v C

y x y C

+ + + =

= −+

+ = − +

+ =

+ =

∫ ∫

69. ( )2

2

3 4

3 12

3 12

y x dx x dy

dyx y xdx

y y xx

− = −

= − +

′ + =

Integrating factor: ( )3 3 ln 3x dx xe e x∫ = =

( )3 3 3 4

3 4 5

23

3 12 12

12125

125

y x x y x x xx

yx x dx x C

Cy xx

′ + = =

= = +

= +

∫

70. ( )( )2 1 0yx dx y e x dy+ + + =


( )

( )( )

2

2 21

2 2

11 1ln 12 2

ln 1 2

y

y

y

x dx y e dyx

x y e C

x y e C

= − ++

+ = − − +

+ + + =

∫ ∫

71. False. The equation contains .y

72. True. ( ) 0xy x e y′ + − = is linear.

Review Exercises for Chapter 6

1.

( ) ( )3 2

2 3 3

, 3

2 4 2 3 4 10 .

y x y x

xy y x x x x

′= =

′ + = + =

Yes, it is a solution.

2.

( )

2 sin 24 cos 2

8 sin 216 cos 2

8 16 cos 2 8 2 sin 2 0

y xy xy xy x

y y x x

=

′ =

′′ = −

′′′ = −

′′′ − = − − ≠

Not a solution

3.

( )

2

32

4 7

44 7 73

dy xdx

xy x dx x C

= +

= + = + +∫

4.

( )

3

3 4 2

3 8

33 8 44

dy x xdx

y x x dx x x C

= −

= − = − +∫

5. cos 2

1cos 2 sin 22

dy xdx

y x dx x C

=

= = +∫

6. 2 sin

2 sin 2 cos

dy xdx

y x dx x C

=

= = − +∫

7. 5

5

dy x xdx

y x x dx

= −

= −∫

Let 5, , 5u x du dx x u= − = = +

( )

( )

( ) ( )

( ) ( )

( ) ( )

3 2 1 2

5 2 3 2

5 2 3 2

3 2

3 2

5

5

2 105 32 105 55 31 5 6 5 50

151 5 6 20

15

y u u du

u u du

u u C

x x C

x x C

x x C

= +

= +

= + +

= − + − +

= − ⎡ − + ⎤ +⎣ ⎦

= − + +

∫∫



8. 2 7

2 7

dy x xdx

y x x dx

= −

= −∫

Let 7, , 7:u x du dx x u= − = = +

( )

( ) ( )

( ) ( )

1 2

5 2 3 2

5 2 3 2

3 2

2 7

4 285 34 287 75 34 7 3 14

15

y u u du

u u C

x x C

x x C

= +

= + +

= − + − +

= − + +

∫

9. 2

2 2

x

x x

dy edx

y e dx e C

−

− −

=

= = − +∫

10. 3

3 3

3

3 9

x

x x

dy edx

y e dx e C

−

− −

=

= = − +∫

11. 2dy x ydx

= −

12. sin4

dy yxdx

π⎛ ⎞= ⎜ ⎟⎝ ⎠

13. ( )3 , 2, 1y x′ = −

(a) and (b)

14. ( )22 , 0, 2y x x′ = −

(a) and (b)

15. ( )21 14 3 , 0, 3y x x′ = −

(a) and (b)

16. ( )4 , 1, 1y y x′ = + −

(a) and (b)

17. ( )2 , 0, 14

xyyx

′ =+

(a) and (b)

18. ( )2 , 0, 21

yyx

′ = −+

(a) and (b)

x –4 −2 0 2 4 8

y 2 0 4 4 6 8

dy dx –10 –4 –4 0 2 8

x –4 −2 0 2 4 8

y 2 0 4 4 6 8

dy dx –4 0 0 0 –4 0

x

y(2, 1)

4

8−2

y

x

(0, 2)

3−3

5

−1

−4 4

−4

4

x

y

(0, 3)

y

x3−3

2

−4

(−1, 1)

4

−4

4

x

y

(0, 1)

−4

y

x

(0, −2)

6−6

−10

2



19.

( )2

8

8 82

dy xdx

xy x dx x C

= −

= − = − +∫

20.

1

1

8

8

ln 8

8

8

x C x

x

dy ydx

dy dxy

y x C

y e Ce

y Ce

+

= +

=+

+ = +

+ = =

= − +

∫ ∫

21. ( )

( )

( )

2

2

1

3

3

3

13

13

dy ydx

y dy dx

y x C

yx C

yx C

−

−

= +

+ =

− + = +

−+ =

+

= − −+

∫ ∫

22.

( )

1 2

1 21

11 2

2

10

10

2 10

52

5

dy ydx

y dy dx

y x C

Cy x C C

y x C

−

=

=

= +

⎛ ⎞= + =⎜ ⎟⎝ ⎠

= +

∫ ∫

23.

( )

( )

( )( )

1

22

2 0

2

12

1 212

ln 2 ln 2

22

xx

x y xy

dyx xydx

xdy dxy x

dy dxy x

y x x C

Cey Ce xx

−

′+ − =

+ =

=+

⎛ ⎞= −⎜ ⎟+⎝ ⎠

= − + +

= + =+

24. ( )

( )

1

1 0

1

1

ln lnx

xy x y

dyx x ydxdy x dxy x

y x x C

y Cxe

′ − + =

= +

+=

= + +

=

∫ ∫

25. kty Ce=

( )( ) ( )

( )

5

5

3 34 4

34

203

2015 3

0, :

5, 5 : 5

ln

k

k

C

e

e

k

=

=

=

=

( )ln 20 3 5 0.3793 34 4

t ty e e⎡ ⎤⎣ ⎦= ≈

26. kty Ce=

( )( ) ( )

( )

2 2

4 2 4 2

2

3 3 32 2 2

3 32 2

10 1013 2 3

2, :

4, 5 : 5

ln

k k

k k k k

k

Ce C e

Ce e e e

e k

−

−

= ⇒ =

= = =

= ⇒ =

So, ( ) ( ) ( )2 1 2 ln 10 33 3 3 92 2 10 20.C e−= = =

Finally, ( )1 2 ln 10 3920 .ty e=

27. kty Ce=

( )0, 5 : 5C =

51 15, : 56 6

1 1 ln 30ln5 30 5

ke

k

⎛ ⎞ =⎜ ⎟⎝ ⎠

−⎛ ⎞= =⎜ ⎟⎝ ⎠

[ ]ln30 5 0.6805 5t ty e e− −= ≈

28. kty Ce=

( )( ) ( )

( )

6 6 5

1 25 9

1, 9 : 9 9

6, 2 : 2 2 9 9

ln 0.3008

k k

k k k k

Ce C e

Ce e e e

k

−

−

= ⇒ =

= ⇒ = =

= ≈ −

So, ( ) ( ) 1 51 5 ln 2 9 299 9 12.15864.C e

−−= = ≈

Finally, 0.300812.1586 .ty e−≈

29. ( ), 0 30dP kp Pdh

= =

( )( )

( )

( ) ( )

( ) ( )

18,000

ln 2 18,000

35,000 ln 2 18,000

30

18,000 30 15

ln 1 2 ln 218,000 18,000

30

35,000 30 7.79 inches

kh

k

h

P h e

P e

k

P h e

P e

−

−

=

= =

−= =

=

= ≈



30. ( )1599

15

7.5 15

kt kt

k

y Ce e

e

= =

=

( )1 11599 2ln 0.000433k = ≈ −

When ( )0.000433 750750, 15 10.84 .t y e g−= = ≈

31. k tS Ce=

(a)

( ) 1.7918

16

16

5 when 1

5

lim 30

5 30

ln 1.7918

30 30

k

k tt

k

t

S t

Ce

Ce C

e

k

S e

→∞

−

= =

=

= =

=

= ≈ −

= ≈

(b) When 5, 20.9646t S= ≈ which is 20,965 units.

(c)

32. ( )25 1 ktS e= −

(a) ( )( ) ( )( )

1

0.1744

4 21 2125 25 254 25 1 1 ln 0.1744

25 1

k k k

t

e e e k

S e−

= − ⇒ − = ⇒ = ⇒ = ≈ −

= −

(b) 25,000 units ( )lim 25t

S→∞

=

(c) When 5, 14.545t S= ≈ which is 14,545 units.

(d)

33. 0.0185

0.0185

0.0185

2

2ln 2 0.0185

ln 2 37.5 years0.0185

t

t

t

P Ce

C Ce

et

t

=

=

=

=

= ≈

34. (a)

1

0.012

0.012 , 50

10.012

1 ln0.012

s

dy y sdsdy dsy

y s C

y Ce−

= − >

−=

−= +

=

∫ ∫

When ( )0.012 50 0.650, 28 28s y Ce C e−= = = ⇒ =

0.6 0.01228 , 50.sy e s−= >

(b)

Speed (s) 50 55 60 65 70

Miles per Gallon (y) 28 26.4 24.8 23.4 22.0

00

40

30

0 80

25



35. 5

5

7 7

7 7 ln5

dy x xdx x x

xy x dx x Cx

4

4

+= = +

⎛ ⎞= + = + +⎜ ⎟⎝ ⎠∫

36.

( )

2

2

2 2

2 2

2

11 2

1 2 11 ln 12

x

x

x x

x x

x

dy edx e

e edy dx dxe e

y e C

−

−

− −

− −

−

=+

−= = −

+ +

= − + +

∫ ∫ ∫

37.

21

28 1

28

16 0

16

1 16

ln 8

x C

x

y xydy xydx

dy x dxy

y x C

e y

y Ce

+

′ − =

=

=

= +

=

=

∫ ∫

38.

( )

1

1

sin 0

sin

sin

cos1

cos

1ln ln coscos

y

y

y

y

y

y e xdy e xdx

e dy x dx

e x C

e C Cx C

y x Cx C

−

−

′ − =

=

=

− = − +

= = −+

= = − ++

∫ ∫

39. 2 2

2dy x ydx xy

+= (homogeneous differential equation)

( )2 2 2 0x y dx xy dy+ − =


( ) ( )( )

( )( )

( )

( )

2 2 2

2 2 2 2 2 3

2 2 2 3

2

2

2 21

2

22 2 2

12 2 2 2

2 0

2 2 0

2

1 2

21

ln ln 1 ln 1 ln

1 1

1 or


x v x x v dx x v dv

x x v dx x v dv

v dx xv dv

dx v dvx vx v C v C

C C Cxxv x yy x

Cx xCx y x y

+ − + =

+ − − =

− =

− =

=−

= − − + = − − +

= = =− −−

= =− −

∫ ∫



40. ( )3 x ydydx x

+= (homogeneous differential equation)

( )3 0x y dx x dy+ − =


( ) ( )( )

( )

( )

( )

( )

( )

2

1 21 2

1 22

2

3

3

3 0

3 2 0

3 2

1 13 2

1ln ln 3 2 ln 3 2 ln2

3 2

3 2 3 2

3 2 3 2

32

x vx dx x x dv v dx

x vx dx x dv

v dx x dv

dx dvx v

x v C v C

x C v

yx C v Cx

x C x y Cx Cy

x CxyC

+ − + =

+ − =

+ =

=+

= + + = + +

= +

⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= + = +

−=

∫ ∫

41. 31 2

21 2

2

36

y C x C x

y C C xy C x

= +

′ = +

′′ =

( ) ( ) ( )2 2 2 3

2 1 2 1 2

3 3 32 1 2 1 2

3 3 6 3 3 3

6 3 9 3 3 0

x y xy y x C x x C C x C x C x

C x C x C x C x C x

′′ ′− + = − + + +

= − − + + =

1 2 1 22, 0: 0 2 8 4x y C C C C= = = + ⇒ = −

( )1 2

2 2 2 2 1

3

12

12

2, 4: 4 12

4 4 12 8 , 2

2

x y C C

C C C C C

y x x

′= = = +

= − + = ⇒ = = −

= − +

42. 9.8dv kvdt

= −

(a)

1

2

23

3

9.81 ln 9.8

ln 9.8

9.81 9.8

kt C kt

kt

dv dtkv

kv t Ck

kv kt C

kv e C e

v C ek

+

=−

− = +

− = +

− = =

⎡ ⎤= +⎣ ⎦

∫ ∫

At ( )

( )

0 3 3 0

0

10, 9.8 9.8

1 9.8 9.8 .kt

t v C C kvk

v kv ek

= = + ⇒ = −

⎡ ⎤= + −⎣ ⎦

Note that 0k < because the object is moving downward.

(b) ( ) 9.8limt

v tk→∞

=



(c) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )( )

0 0 02

0 0 02 2

0 0 0 0 02 2 2

1 1 1 9.8 19.8 9.8 9.8 9.8 9.8

1 10 9.8 9.8

9.8 1 1 9.8 19.8 9.8 9.8 1

kt kt kt

kt kt

ts t kv e dt t kv e C kv e Ck k k k k

s kv C C s kvk k

t ts t kv e s kv kv e sk k k k k

⎡ ⎤⎡ ⎤= + − = + − + = + − +⎢ ⎥⎣ ⎦ ⎣ ⎦

= − + ⇒ = − −

= + − + − − = + − − +

∫

43. 4dy xdx y

−=

2

21

2 2

4

22

4 ellipses

y dy x dx

y x C

x y C

= −

= − +

+ =

∫ ∫

44. 3 2dy ydx

= −

1

1

22

22

2

2 31 ln 2 32

ln 2 3 2 2

2 3

2 332

x

x

x

dy dxy

y x C

y x C

y C e

y C e

y Ce

−

−

−

= −−

− = − +

− = − +

− =

= +

= +

∫ ∫

45. ( ) 0.555250

1 34 tP te−

=+

(a) 0.55k =

(b) 5250L =

(c) ( ) 52500 1501 44

P = =+

(d) 0.55

0.55

0.55

525026251 34

1 34 2134

1 1ln 6.41 yr0.55 34

t

t

t

ee

e

t

−

−

−

=+

+ =

=

− ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

(e) 0.55 15250

dP PPdt

⎛ ⎞= −⎜ ⎟⎝ ⎠

46. ( ) 0.154800

1 14 tP te−

=+

(a) 0.15k =

(b) 4800L =

(c) ( ) 48000 3201 14

P = =+

(d) 0.15

0.15

480024001 14

14 1

1 1ln 17.59 yr0.15 14

t

t

ee

t

−

−

=+

=

⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

(e) 0.15 14800

dP PPdt

⎛ ⎞= −⎜ ⎟⎝ ⎠

47. ( )1 , 0, 880

dy yydt

⎛ ⎞= −⎜ ⎟⎝ ⎠

( )

1, 8080

1 1800 8: 8 9

1

kt t

k LLybe be

y bb

− −

= =

= =+ +

= = ⇒ =+

Solution: 801 9 ty

e−=

+

x

y

4

−4

−4 4

4x

y



48. ( )1.76 1 , 0, 38

dy yydt

⎛ ⎞= −⎜ ⎟⎝ ⎠

( )

1.76

1.76, 88

1 18 50 3: 3

1 3

kt t

k LLybe be

y bb

− −

= =

= =+ +

= = ⇒ =+

Solution: 1.76

8513

ty

e−=

⎛ ⎞+ ⎜ ⎟⎝ ⎠

49. (a)

( ) ( )

( )

( )

0.553

20,400, 0 1200, 1 2000

20,4001

20,4000 1200 16120,4001 2000

1 1646165

23 40ln ln 0.55340 23

20,4001 16

kt

k

k

t

L y y

ybe

y bb

ye

e

k

ye

−

−

−

−

= = =

=+

= = ⇒ =+

= =+

=

= − = ≈

=+

(b) ( )8 17,118 trouty ≈

(c) 0.55320,40010,000 4.94 yr

1 16 t te−

= ⇒ ≈+

50. ( )0.553 1 , 0 120020,400

dy yy ydt

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

Use Euler's method with 1.h =

Euler's method gives ( )8 16,170 trout.y ≈

51.

( )

( )

10

1, ( ) 10dx x

y y

P x Q x

u x e e− −

′ − =

= − =

∫= =

( )

1 10

10

10

xx

x x

x

y e dxee e C

Ce

−−

−

=

= − +

= − +

∫

52. 4 1

4

x x

x

e y e y

y y e−′ + =

′ + =

( )

( ) 4 4

4, ( ) x

dx x

P x Q x e

u x e e

−= =

∫= =

4 4 3 441 1 1

3 3x x x x x x

xy e e dx e e C e Cee

− − − −⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∫

53.

4

41 14 4

x y

x

y e y

y y e

′ = +

′ − =

( ) ( )

( ) ( ) ( )

4

1 4 1 4

1 1,4 4

x

dx x

P x Q x e

u x e e− −

= − =

∫= =

( )( )

( )

1 441 4

1 4

4 4

1 14

14

14

xxx

x

x x

y e e dxe

e x C

xe Ce

−−

=

⎛ ⎞= +⎜ ⎟⎝ ⎠

= +

∫

54. 2 25 1dy y

dx x x− =

( ) ( )

( ) ( )2 2

25 5

5 ,

x dx x

P x Q xx x

u x e e−

1= − =

∫= =

5 5 55 2 51 1 1 1 1

5 5x x x

x xy e dx e C Cee x e

−⎛ ⎞= = − + = − +⎜ ⎟⎝ ⎠∫

55. ( )2 1x y y′− + =

1 12 2

dy ydx x x

+ =− −

( ) ( )

( ) ( )1 2 ln 2

1 ,2 2

2x dx x

P x Q xx x

u x e e x− −

1= =

− −∫= = = −

( ) ( )1 1 122 2 2

y x dx x Cx x x

⎛ ⎞= − = +⎜ ⎟− − −⎝ ⎠∫

t 0 2 4 6 8

Exact 1200 3241 7414 12,915 17,117

Euler 1200 2743 5853 10,869 16,170



56. ( ) ( )23 2 2 3x y y x′+ + = +

( )2 2 33

dy y xdx x

+ = ++

( ) ( ) ( )

( ) ( ) ( ) ( )22 3 2 ln 3

2 , 2 33

3x dx x

P x Q x xx

u x e e x+ +

= = ++∫= = = +

( )( )( )

( )( )

( )( )

22

4

2

2

2

1 2 3 33

3123

32 3

y x x dxx

xC

x

x Cx

= + ++

⎡ ⎤+⎢ ⎥= +⎢ ⎥+ ⎣ ⎦

+= +

+

∫

57. ( )3 sin 2 0

3 sin 2

y x dx dy

y y x

+ − =

′ − =

Integrating factor: 3 3dx xe e− −∫ =

( )( )

3 3

3

3

113

113

sin2

3 sin 2 2 cos 2

3 sin 2 2 cos 2

x x

x

x

ye e x dx

e x x C

y x x Ce

− −

−

=

= − − +

= − + +

∫

58. ( )( )

tan 2

tan 2

x

x

dy y x e dx

dy x y edx

= +

− =

Integrating factor: tan ln cos cosx dx xe e x−∫ = =

( )

( )

cos 2 cos cos sin

1 tan sec

x x

x

y x e x dx e x x C

y e x C x

= = + +

= + +

∫

59. 55 xy y e′ + =

Integrating factor: 5 5dx xe e∫ =

5 10 10

5 5

110

110

x x x

x x

ye e dx e C

y e Ce−

= = +

= +

∫

60. 3ay y bxx

⎛ ⎞′ − =⎜ ⎟⎝ ⎠

Integrating factor: ( ) lna x dx a x ae e x− − −∫ = =

( )3 4

4

4

4

a a a

a

byx bx x dx x Ca

bxy Cxa

− − −= = +−

= +−

∫

61. 2 Bernoulli equationy y xy′ + =

1 2 1 22, let , .n z y y z y y− − −′ ′= = = = −

( ) ( )2 2

Linear equation

y y y y x

z z x

− −′− + − = −

′ − = −

( )

1

1

11

1

dx x

x x x xx

x

x

u x e e

z xe dx e xe e Ce

y x Ce

yx Ce

− −

− − −−

−

∫= =

⎡ ⎤= − = + +⎣ ⎦

= + +

=+ +

∫

62. 22 Bernoulli equationy xy xy′ + =

1 2 1 22, let , .n z y y z y y− − −′ ′= = = = −

( ) ( )2 22

2 Linear equation

y y xy y x

z xz x

− −′− + − = −

′ − = −

( ) 22x dx xu x e e− −∫= =

( ) 2 2 22

2

221

1 12

1 12

1 21 12

x x xx

x

xx

z x e dx e e Ce

Cey

yC eCe

− −

−

⎛ ⎞= − = +⎜ ⎟⎝ ⎠

= +

= =++

∫

63. 3

21 Bernoulli equationyy yx x

′ + =

1 3 2 33, let , 2 .n z y y z y y− − −′ ′= = = = −

( ) ( )3 3

2

2

1 22 2

2 2 Linear equation

y y y yx x

z zx x

− − −′− + − =

−′ − =

( ) ( )2 2 ln 2x dx xu x e e x− − −∫= = =

( )

32 2

2 2

2

1 2 23

1 23

xz x dx x Cx x

Cxy x

−−

−

2

⎛ ⎞−= = +⎜ ⎟

⎝ ⎠

= +

∫



64. 2

2

1 2 1 2

1 Bernoulli Equation

2, let , .

xy y xy

y y yx

n z y y z y y− − −

′ + =

′ + =

′ ′= = = = −

( ) ( )2 2 2 21

1 1 Linear equation

y y y y y yx

z zx

− − −′− + − = −

′ − = −

( ) ( )1 1x dxu x ex

−∫= =

1 ln

1 ln

1ln

z x dx x x Cx

x x Cxy

yCx x x

⎡ ⎤= − = − +⎣ ⎦

= − +

=−

∫

65. Answers will vary. Sample answer: ( )2 23 2 0x y dx xy dy+ − =

Solution: Let , .y vx dy x dv v dx= = +

( ) ( )( )

( )( )

( )

( )

2 2 2

2 2 2 3

2

2

21

22

2

3 2 2

3 2 0

2 0

1 2

21

ln ln 1

1 1


x v x dx x v dv

v dx xv dv

dx v dvx vx v C

yx C v Cx

x C x y

+ − + =

+ − =

+ =

=+

= + +

⎛ ⎞= + = +⎜ ⎟

⎝ ⎠

= +

∫ ∫

66. Answers will vary. Sample answer: 140yy y⎛ ⎞′ = −⎜ ⎟

⎝ ⎠

Solution: 1, 40k L= =

401 1kt t

Lybe be− −= =

+ +

67. Answers will vary.

Sample answer: 3 2

3

2 12 1

x y x y

y yx x

′ + =

′ + =

( ) ( )

( )

2 2

22 3 2

1 1 1 ln

x dxu x e x

y x dx x Cx x x

∫= =

⎡ ⎤= = +⎣ ⎦∫

68. Answers will vary. Sample answer: 1y xy xy−′ + =

Solution: 221, , , x dx xn Q x P x e e∫= − = = =

2 2 22

22

2

1

x x x

x

y e xe dx e C

y Ce−

= = +

= +

∫

Problem Solving for Chapter 6 649


Problem Solving for Chapter 6

1. (a)

( )

1.01

1.01

0.01

1

0.01

0.01

100

0.011 0.01

10.01

10.01

dy ydt

y dy dt

y t C

t Cy

yC t

yC t

−

−

=

=

= +−

= − +

=−

=−

∫ ∫

( ) 10010 1: 1 1y C

C= = ⇒ =

So, ( )100

1 .1 0.01

yt

=−

For 100, lim .t T

T y−→

= = ∞

(b) ( )

( )

1

1

11

y dy k dt

y kt C

y kt C

yC kt

ε

ε

ε

ε

εε

ε

− +

−

−

=

= +−

= − +

=−

∫ ∫

( ) 10 1

0 0

1 1 10y y C CC y y

εε

ε

⎛ ⎞= = ⇒ = ⇒ = ⎜ ⎟

⎝ ⎠

So, 1

0

1 .1

y

kty

ε

ε ε

=⎛ ⎞

−⎜ ⎟⎝ ⎠

For 0

1 , .t yy kεε

→ → ∞

2. Because ( )20 ,dy k ydt

= −

120

ln 20

20.kt

dy k dty

y kt C

y Ce

=−

− = +

= +

∫ ∫

When 0, 72.t y= = Therefore, 52.C =

When 1, 48.t y= = Therefore, 28 748 52 20, ,52 13

k ke e= + = = and 7ln .13

k = So,

( )ln 7 1352 20.ty e⎡ ⎤⎣ ⎦= +

When ( )5 ln 7 135, 52 20 22.35 F.t y e= = + ≈ °

3. (a) ( )1dS k S L Sdt

= −

1 kt

LSCe−

=+

is a solution because

( ) ( )

( )

( )

2

2

1 1

1

1

1 1

1 1

, where .

kt kt

kt

kt

kt

kt kt

kt kt

dS L Ce Ckedt

LC ke

Ce

k L C LeL Ce Cek L LLL Ce Ce

kk S L S kL

−− −

−

−

−

− −

− −

= − + −

=+

⎛ ⎞= ⋅⎜ ⎟ + +⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

= − =

100.L = Also, 10S = when 0 9.t C= ⇒ =

And, 20S = when 41 ln .9

t k= ⇒ = −

Particular Solution: ( ) 0.8109ln 4 9100 100

1 91 9 ttSee −= =

++

(b) ( )

( )

( )

1

2

12

1

100

100

100 2

0 when 50 or 0.

dS k S Sdt

d S dS dSk S Sdt dt dt

dSk Sdt

dSSdt

= −

⎡ ⎤⎛ ⎞= − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= −

= = =

Choosing 50,S = you have:

( )

( )

( )( )

ln 4 9

ln 4 9

100501 9

2 1 9ln 1 9ln 4 9

2.7 months

t

tee

t

t

=+

= +

=

≈

(This is the point of inflection.) (c)

(d)

(e) Sales will decrease toward the line .S L=

00

10

125

t1 2 3 4

140

120

100

80

60

40

20

S



4. (a)

[ ][ ] 1

ln

ln ln

ln ln ln

ln kt

ktCe

ktCe

dy Lk ydt y

dy k dty L y

L y kt C

L CeyL ey

y Le

−

−

−−

⎛ ⎞= ⎜ ⎟

⎝ ⎠

=−

− = − +

=

=

= (b)

(c) As , ,t y L→ ∞ → the carrying capacity.

(d) 0 500 5000 10 ln 10C Cy e e C−= = ⇒ = ⇒ =

( )

2

2 2

2

ln

1ln

ln 1 ln ln 1

dy Lk ydt y

d y L dy L dyk kydt y dt L y y dt

dy L L Lk k ydt y y y

⎛ ⎞= ⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞−= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

So, 2

2 0d ydt

= when

ln 1 .L L Le yy y e

⎛ ⎞= ⇒ = ⇒ =⎜ ⎟

⎝ ⎠

5000 1839.4Lye e

= = ≈ and 41.7.t ≈

The graph is concave upward on ( )0, 41.7 and

downward on ( )41.7, .∞

5. Let 1 ln .2

bu k tk

⎛ ⎞= −⎜ ⎟⎝ ⎠

( )( )

4

2

ln2 ln

21 tanh 11

u

u u u

k t b ku b kt kt

e eue e e

e e e e be

−

− −

− −− − −

−+ = + =

+ +

= = =

Finally,

[ ]1 1 ln1 tanh 1 tanh2 2 2

22 1

.1

kt

kt

b LL k t uk

Lbe

Lbe

−

−

⎡ ⎤⎛ ⎞⎛ ⎞+ − = +⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

=+

=+

Notice the graph of the logistics function is just a shift of the graph of the hyperbolic tangent. (See section 5.8.)

6. ( ) ( ) ( ) ( )?

f x g x f x g x′ ′ ′⎡ ⎤ =⎣ ⎦

(a) Let ( ) ( ), 1,g x x g x′= = then

( ) ( )( ) ( ) ( )

( ) ( )

( )

( )

1

1

ln ln 1

11

f x x f x

f x x f x f x

df x f xdx

df dxf x

f x x

f xx

′ ′⎡ ⎤ =⎣ ⎦′ ′+ =

− = −

=−

= − −

=−

∫ ∫

(b) ( )

( )

ln

g dxg g

fg f g

f g fg f g

f g g fg

f gf g g

gf dxg g

f e′

′ −

′ ′ ′=

′ ′ ′ ′+ =

′ ′ ′− = −

′ ′=

′ −′

=′ −

∫=

∫

(c) If ( ) ,xg x e= then ( ) ( ) 0x xg x g x e e′ − = − =

Therefore, no f can exist.

2000

00 500

7000

00 500



7. 21

1232

k

g

π⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )

( )

22

22 2

6 36 Equation of tank

36 6 12

x y

x y y y

+ − =

= − − = −

Area of cross section: ( ) ( )212A h h h π= −

( )

( )

( )( )

( )

2

2 1 2

3 2 1 2

5 2 3 2

3 2

2

112 64144112

1818 216

36 1445

36 7205

dhA h k ghdtdhh h hdtdhh h hdt

h h dh dt

h h t C

h h t C

π π

= −

− = −

− = −

− =

− = +

− = +

∫ ∫

When 6, 0h t= = and ( )3 26 504 1481.45.5

C = − ≈ −

The tank is completely drained when 0 1481.45 sec 24 min, 41 sech t= ⇒ = ≈

8. (a) ( )

( )

2

1 22 2

1

1

2

64

8 8,

2

2 18 at 0, 18

dhA h k ghdt

dhr k hdt

k kh dh dt C dt Cr r

h Ct C

C t h

π

π π−

= −

= −

−= = − =

= − +

= = =

So, 2 6 2.h Ct= − +

At ( )30 60 1800, 12:t h= = =

2 12 1800 6 2

6 2 4 3 0.0008651800

C

C

= − +

−= ≈

So, 2 0.000865 6 2.h t= − +

( )

6 200.0008659809.1seconds 2 h, 43 min, 29 sec

h t= ⇒ =

≈

(b) ( )3600 sec 2 0.000865 3600 6 2

7.21 ft

t h

h

= ⇒ = − +

⇒ ≈

9.

( )

1 2

2

64 836

1288

2288

dhA h k ghdtdh hdt

h dh dt

th C

ππ

−

= −

−=

−=

−= +

∫ ∫

20: 2 20 4 5

2 4 5288

h Cth

= = =

−= +

( )0 4 5 288

2575.95 sec 42 min, 56 sec

h t= ⇒ =

≈ ≈

h

6 ft

x2 + (y − 6)2 = 36

x

y

h

r

18 ft



10. Let the radio receiver be located at ( )0 , 0 .x The tangent line to 2y x x= − joins ( )1, 1− and ( )0 , 0 .x

(a) If ( ),x y is the point of tangency on 2,y x x= − then

2

2 2

2

2

1 11 21 1

2 1 2 1

2 2 0

2 4 8 1 32

3 3 5.

y x xxx x

x x x x x

x x

x

y x x

− − −− = =

+ +− + − = − −

+ − =

⎛ ⎞− ± += = − +⎜ ⎟⎜ ⎟⎝ ⎠

= − = −

Then

( )( ) ( )0

0 0

0

1 0 1 3 3 5 6 3 31 1 1 3 3

3 1 6 3 3 6 3 3 6 3 3

4 3 6 1.155.6 3 3

x

x x

x

− − + −= =

− − − + − −

= + − = − + −

−= ≈

−

(b) Now let the transmitter be located at ( )1, .h−

( )( )

2

2 2

2

2

1 21 1

2 1 2

2 1 0

2 4 4 11 2

23 2 4

y h x x hxx x

x x x x x h

x x h

hx h

y x x h h

− − −− = =

+ +− + − = − −

+ − − =

− ± + += = − + +

= − = + − −

Then, ( )

( )0

0

0

3 2 40 2 4 3 21 21 1 2

1 22 4 3 2

2 1.2 4 3 2

h h hh h hx hh

x hh h h

h hxh h

− + − −− + − += =

− − − +− − − + +

+ +=

+ − +

+= −

+ − +

(c)

There is a vertical asymptote at 1,4

h = which is the height of the mountain.

xx

x0

−1 1

2

1

y x x= − 2

Transmitter

Radio

( 1, 1)−

y

0.25 3

−2

10



11. 3.5 0.019ds sdt

= −

(a) 3.5 0.019

ds dts

−= −

−∫ ∫

1

2

0.0193

0.0193

0.019

1 ln 3.5 0.0190.019

ln 3.5 0.019 0.019

3.5 0.0190.019 3.5

184.21

t

t

t

s t C

s t C

s C es C es Ce

−

−

−

− = − +

− = − +

− =

− = −

= −

(b)

(c) As 0.019, 0,tt Ce−→ ∞ → and 184.21.s →

12. (a)

1ln

Rt V

dC R dtC V

RC t KV

C Ke−

= −

= − +

=

∫ ∫

Since 0C C= when 0,t = it follows that 0K C= and the function is 0 .Rt VC C e−= (b) Finally, as ,t → ∞ we have

0lim lim 0.Rt Vt t

C C e−→∞ →∞

= =

13. From Exercises 12, you have 0 .Rt VC C e−=

(a) For 2, 0.5,V R= = and 0 0.6,C = you have 0.250.6 tC e−=

(b) For 2, 1.5,V R= = and 0 0.6,C = you have 0.750.6 .tC e−=

14. (a)

( )

( )( ) ( )

1

1

1

1 1

1 ln

1 1

R t V K

R t V K Rt V

dC dtQ RC V

tQ RC KR V

Q RC e

C Q e Q KeR R

− +⎡ ⎤⎣ ⎦

− +⎡ ⎤ −⎣ ⎦

=−

− − = +

− =

= − = −

∫ ∫

Because 0C = when 0,t = it follows that K Q= and you have ( )1 .Rt VQC eR

−= −

(b) As ,t → ∞ the limit of C is .Q R

00 200

400

00

4

0.8

00

4

0.8

chapter 6 differential equations - seneca valley … 6 differential equations section 6.1 slope...

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